Computing $int_0^{infty} dfrac{1}{(1+x^2)^2}dx$ using Plancherel












4












$begingroup$


I want to compute $int_0^{infty} dfrac{1}{(1+x^2)^2}dx$, using Plancherel.



So define $f(x) = dfrac{1}{1+x^2}$ , then $fin L_1cap L_2$ so by Plancherel we know that $||f||_2 =||hat{f}||_2$ where $hat{f}(t) = dfrac{1}{sqrt{2pi}}int_{-infty}^{infty}f(x)e^{-ixt}dx$.



So we have that $$int_0^{infty} dfrac{1}{(1+x^2)^2}dx = ||f||_2^2 = ||hat{f}||_2^2 = int_0^{infty}( dfrac{1}{sqrt{2pi}}int_{0}^{infty}f(x)e^{-ixt}dx)^2dt = $$



$$dfrac{1}{2pi}int_0^infty(int_0^inftydfrac{e^{-ixt}}{1+x^{2}}dx)^2dt$$



This was the guidance in the exercise (using Plancherel) and it seems a bit difficult (or maybe I'm missing something). Is the inner integral easy to compute? If so, how?



Thanks for helping!






real-analysis fourier-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    You might find the formula $mathcal{F}left[(1 + x^2)^{-1}right] = e^{-|xi|}$ useful, which is true up to a normalization.
    $endgroup$
    – T. Bongers
    Dec 2 '18 at 21:32










  • $begingroup$
    Do you know which function has $f$ as the Fourier transform?
    $endgroup$
    – Fakemistake
    Dec 2 '18 at 21:34










  • $begingroup$
    Actually, this question might be close enough to close as duplicate.
    $endgroup$
    – T. Bongers
    Dec 2 '18 at 21:35










  • $begingroup$
    @Fakemistake i think this is the inner integral
    $endgroup$
    – user123
    Dec 2 '18 at 21:39










  • $begingroup$
    @T.Bongers thanks for that link. i will look at it tomorrow and if it solves my question i will delete it.
    $endgroup$
    – user123
    Dec 2 '18 at 21:42
















4












$begingroup$


I want to compute $int_0^{infty} dfrac{1}{(1+x^2)^2}dx$, using Plancherel.



So define $f(x) = dfrac{1}{1+x^2}$ , then $fin L_1cap L_2$ so by Plancherel we know that $||f||_2 =||hat{f}||_2$ where $hat{f}(t) = dfrac{1}{sqrt{2pi}}int_{-infty}^{infty}f(x)e^{-ixt}dx$.



So we have that $$int_0^{infty} dfrac{1}{(1+x^2)^2}dx = ||f||_2^2 = ||hat{f}||_2^2 = int_0^{infty}( dfrac{1}{sqrt{2pi}}int_{0}^{infty}f(x)e^{-ixt}dx)^2dt = $$



$$dfrac{1}{2pi}int_0^infty(int_0^inftydfrac{e^{-ixt}}{1+x^{2}}dx)^2dt$$



This was the guidance in the exercise (using Plancherel) and it seems a bit difficult (or maybe I'm missing something). Is the inner integral easy to compute? If so, how?



Thanks for helping!






real-analysis fourier-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    You might find the formula $mathcal{F}left[(1 + x^2)^{-1}right] = e^{-|xi|}$ useful, which is true up to a normalization.
    $endgroup$
    – T. Bongers
    Dec 2 '18 at 21:32










  • $begingroup$
    Do you know which function has $f$ as the Fourier transform?
    $endgroup$
    – Fakemistake
    Dec 2 '18 at 21:34










  • $begingroup$
    Actually, this question might be close enough to close as duplicate.
    $endgroup$
    – T. Bongers
    Dec 2 '18 at 21:35










  • $begingroup$
    @Fakemistake i think this is the inner integral
    $endgroup$
    – user123
    Dec 2 '18 at 21:39










  • $begingroup$
    @T.Bongers thanks for that link. i will look at it tomorrow and if it solves my question i will delete it.
    $endgroup$
    – user123
    Dec 2 '18 at 21:42














4












4








4





$begingroup$


I want to compute $int_0^{infty} dfrac{1}{(1+x^2)^2}dx$, using Plancherel.



So define $f(x) = dfrac{1}{1+x^2}$ , then $fin L_1cap L_2$ so by Plancherel we know that $||f||_2 =||hat{f}||_2$ where $hat{f}(t) = dfrac{1}{sqrt{2pi}}int_{-infty}^{infty}f(x)e^{-ixt}dx$.



So we have that $$int_0^{infty} dfrac{1}{(1+x^2)^2}dx = ||f||_2^2 = ||hat{f}||_2^2 = int_0^{infty}( dfrac{1}{sqrt{2pi}}int_{0}^{infty}f(x)e^{-ixt}dx)^2dt = $$



$$dfrac{1}{2pi}int_0^infty(int_0^inftydfrac{e^{-ixt}}{1+x^{2}}dx)^2dt$$



This was the guidance in the exercise (using Plancherel) and it seems a bit difficult (or maybe I'm missing something). Is the inner integral easy to compute? If so, how?



Thanks for helping!






real-analysis fourier-analysis







share|cite|improve this question











$endgroup$




I want to compute $int_0^{infty} dfrac{1}{(1+x^2)^2}dx$, using Plancherel.



So define $f(x) = dfrac{1}{1+x^2}$ , then $fin L_1cap L_2$ so by Plancherel we know that $||f||_2 =||hat{f}||_2$ where $hat{f}(t) = dfrac{1}{sqrt{2pi}}int_{-infty}^{infty}f(x)e^{-ixt}dx$.



So we have that $$int_0^{infty} dfrac{1}{(1+x^2)^2}dx = ||f||_2^2 = ||hat{f}||_2^2 = int_0^{infty}( dfrac{1}{sqrt{2pi}}int_{0}^{infty}f(x)e^{-ixt}dx)^2dt = $$



$$dfrac{1}{2pi}int_0^infty(int_0^inftydfrac{e^{-ixt}}{1+x^{2}}dx)^2dt$$



This was the guidance in the exercise (using Plancherel) and it seems a bit difficult (or maybe I'm missing something). Is the inner integral easy to compute? If so, how?



Thanks for helping!






real-analysis fourier-analysis




real-analysis fourier-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 0:03









Andrews

3901317




3901317










asked Dec 2 '18 at 21:30









user123user123

1,306316




1,306316












  • $begingroup$
    You might find the formula $mathcal{F}left[(1 + x^2)^{-1}right] = e^{-|xi|}$ useful, which is true up to a normalization.
    $endgroup$
    – T. Bongers
    Dec 2 '18 at 21:32










  • $begingroup$
    Do you know which function has $f$ as the Fourier transform?
    $endgroup$
    – Fakemistake
    Dec 2 '18 at 21:34










  • $begingroup$
    Actually, this question might be close enough to close as duplicate.
    $endgroup$
    – T. Bongers
    Dec 2 '18 at 21:35










  • $begingroup$
    @Fakemistake i think this is the inner integral
    $endgroup$
    – user123
    Dec 2 '18 at 21:39










  • $begingroup$
    @T.Bongers thanks for that link. i will look at it tomorrow and if it solves my question i will delete it.
    $endgroup$
    – user123
    Dec 2 '18 at 21:42


















  • $begingroup$
    You might find the formula $mathcal{F}left[(1 + x^2)^{-1}right] = e^{-|xi|}$ useful, which is true up to a normalization.
    $endgroup$
    – T. Bongers
    Dec 2 '18 at 21:32










  • $begingroup$
    Do you know which function has $f$ as the Fourier transform?
    $endgroup$
    – Fakemistake
    Dec 2 '18 at 21:34










  • $begingroup$
    Actually, this question might be close enough to close as duplicate.
    $endgroup$
    – T. Bongers
    Dec 2 '18 at 21:35










  • $begingroup$
    @Fakemistake i think this is the inner integral
    $endgroup$
    – user123
    Dec 2 '18 at 21:39










  • $begingroup$
    @T.Bongers thanks for that link. i will look at it tomorrow and if it solves my question i will delete it.
    $endgroup$
    – user123
    Dec 2 '18 at 21:42
















$begingroup$
You might find the formula $mathcal{F}left[(1 + x^2)^{-1}right] = e^{-|xi|}$ useful, which is true up to a normalization.
$endgroup$
– T. Bongers
Dec 2 '18 at 21:32




$begingroup$
You might find the formula $mathcal{F}left[(1 + x^2)^{-1}right] = e^{-|xi|}$ useful, which is true up to a normalization.
$endgroup$
– T. Bongers
Dec 2 '18 at 21:32












$begingroup$
Do you know which function has $f$ as the Fourier transform?
$endgroup$
– Fakemistake
Dec 2 '18 at 21:34




$begingroup$
Do you know which function has $f$ as the Fourier transform?
$endgroup$
– Fakemistake
Dec 2 '18 at 21:34












$begingroup$
Actually, this question might be close enough to close as duplicate.
$endgroup$
– T. Bongers
Dec 2 '18 at 21:35




$begingroup$
Actually, this question might be close enough to close as duplicate.
$endgroup$
– T. Bongers
Dec 2 '18 at 21:35












$begingroup$
@Fakemistake i think this is the inner integral
$endgroup$
– user123
Dec 2 '18 at 21:39




$begingroup$
@Fakemistake i think this is the inner integral
$endgroup$
– user123
Dec 2 '18 at 21:39












$begingroup$
@T.Bongers thanks for that link. i will look at it tomorrow and if it solves my question i will delete it.
$endgroup$
– user123
Dec 2 '18 at 21:42




$begingroup$
@T.Bongers thanks for that link. i will look at it tomorrow and if it solves my question i will delete it.
$endgroup$
– user123
Dec 2 '18 at 21:42










2 Answers
2






active

oldest

votes


















2












$begingroup$

Let me define $f$ as you have defined, namely $f(x) =(1+x^2)^{-1}$. In this case Plancherel doesn't make the evaluation much easier. However, in these kind of exercises they usually assume that you know that
$$hat f(t) = frac{1}{sqrt{2pi} }int_mathbb{R} e^{-itx} f(x) , dx =sqrt{fracpi 2}e^{-|t|}. $$
For example, finding the Fourier transform of that particular $f$ is asked in a previous exercise or it is given as an example in your notes. Anyways you can find a proof in this link.
Let us now use Plancherel to conclude
begin{align}
int_0^infty frac 1 {(1+x^2)^2},dx&=frac 1 2 int_mathbb{R} |f(x) |^2,dx \
&=frac 1 2 int_mathbb{R} |hat f(x) |^2,dx \
&=frac 1 2 int_mathbb{R} fracpi 2e^{-2|t|},dt end{align}
The latter integral is elementary
$$frac 1 2 int_mathbb{R} fracpi 2e^{-2|t|},dt = int_0^inftyfracpi 2e^{-2t},dt = frac pi 4$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    About your comment about the bounds, thanks. I thought because im trying to find the integral from $0$ to $infty$ then i will only look at this ray. So i need to write the $-infty,infty$ bounds everywhere and just devide by 2 because the function is even.
    $endgroup$
    – user123
    Dec 3 '18 at 5:53










  • $begingroup$
    where you wrote "Let us now use Plancherel to conclude i think the second equation it is suppose to be $overline{f}$ (f with hat, not sure what is the code for it.)
    $endgroup$
    – user123
    Dec 3 '18 at 5:56










  • $begingroup$
    @Liad Yes, I made a small correction, thanks! I removed the comment about the bounds since you were aware of that.
    $endgroup$
    – Shashi
    Dec 3 '18 at 6:55










  • $begingroup$
    alright, thanks for the answer, we indeed calculated the fourier transform of $e^{-alpha |x|}$ , i should've seen the connection. anyways, thanks a lot!
    $endgroup$
    – user123
    Dec 3 '18 at 6:58










  • $begingroup$
    @Liad you are welcome! I don't have time to look at it at the moment, but I will if I have time later in the day (if I won't forget it)
    $endgroup$
    – Shashi
    Dec 3 '18 at 7:06



















2












$begingroup$

This isn't the way that you asked for the integral to be computed, but it's a way that it could be computed.



Consider the indefinite integral
$$I_n=intfrac{mathrm{d}x}{(ax^2+b)^n}$$
Integrating by parts with $mathrm{d}v=mathrm{d}x$ gives
$$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{ax^2 mathrm{d}x}{(ax^2+b)^{n+1}}$$
$$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{ax^2+b}{(ax^2+b)^{n+1}}mathrm{d}x-2bnintfrac{mathrm{d}x}{(ax^2+b)^{n+1}}$$
$$I_n=frac{x}{(ax^2+b)^n}+2nI_n-2bnI_{n+1}$$
$$2bnI_{n+1}=frac{x}{(ax^2+b)^n}+(2n-1)I_n$$
$$I_{n+1}=frac{x}{2bn(ax^2+b)^n}+frac{2n-1}{2bn}I_n$$
$$I_n=frac{x}{2b(n-1)(ax^2+b)^{n-1}}+frac{2n-3}{2b(n-1)}I_{n-1}$$
Plugging in $a=1,,b=1,,n=2$, and evaluating form $x=0$ to $x=infty$ gives your integral as
$$frac12int_0^infty frac{mathrm dx}{1+x^2}=fracpi4$$






share|cite|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3023226%2fcomputing-int-0-infty-dfrac11x22dx-using-plancherel%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Let me define $f$ as you have defined, namely $f(x) =(1+x^2)^{-1}$. In this case Plancherel doesn't make the evaluation much easier. However, in these kind of exercises they usually assume that you know that
    $$hat f(t) = frac{1}{sqrt{2pi} }int_mathbb{R} e^{-itx} f(x) , dx =sqrt{fracpi 2}e^{-|t|}. $$
    For example, finding the Fourier transform of that particular $f$ is asked in a previous exercise or it is given as an example in your notes. Anyways you can find a proof in this link.
    Let us now use Plancherel to conclude
    begin{align}
    int_0^infty frac 1 {(1+x^2)^2},dx&=frac 1 2 int_mathbb{R} |f(x) |^2,dx \
    &=frac 1 2 int_mathbb{R} |hat f(x) |^2,dx \
    &=frac 1 2 int_mathbb{R} fracpi 2e^{-2|t|},dt end{align}
    The latter integral is elementary
    $$frac 1 2 int_mathbb{R} fracpi 2e^{-2|t|},dt = int_0^inftyfracpi 2e^{-2t},dt = frac pi 4$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      About your comment about the bounds, thanks. I thought because im trying to find the integral from $0$ to $infty$ then i will only look at this ray. So i need to write the $-infty,infty$ bounds everywhere and just devide by 2 because the function is even.
      $endgroup$
      – user123
      Dec 3 '18 at 5:53










    • $begingroup$
      where you wrote "Let us now use Plancherel to conclude i think the second equation it is suppose to be $overline{f}$ (f with hat, not sure what is the code for it.)
      $endgroup$
      – user123
      Dec 3 '18 at 5:56










    • $begingroup$
      @Liad Yes, I made a small correction, thanks! I removed the comment about the bounds since you were aware of that.
      $endgroup$
      – Shashi
      Dec 3 '18 at 6:55










    • $begingroup$
      alright, thanks for the answer, we indeed calculated the fourier transform of $e^{-alpha |x|}$ , i should've seen the connection. anyways, thanks a lot!
      $endgroup$
      – user123
      Dec 3 '18 at 6:58










    • $begingroup$
      @Liad you are welcome! I don't have time to look at it at the moment, but I will if I have time later in the day (if I won't forget it)
      $endgroup$
      – Shashi
      Dec 3 '18 at 7:06
















    2












    $begingroup$

    Let me define $f$ as you have defined, namely $f(x) =(1+x^2)^{-1}$. In this case Plancherel doesn't make the evaluation much easier. However, in these kind of exercises they usually assume that you know that
    $$hat f(t) = frac{1}{sqrt{2pi} }int_mathbb{R} e^{-itx} f(x) , dx =sqrt{fracpi 2}e^{-|t|}. $$
    For example, finding the Fourier transform of that particular $f$ is asked in a previous exercise or it is given as an example in your notes. Anyways you can find a proof in this link.
    Let us now use Plancherel to conclude
    begin{align}
    int_0^infty frac 1 {(1+x^2)^2},dx&=frac 1 2 int_mathbb{R} |f(x) |^2,dx \
    &=frac 1 2 int_mathbb{R} |hat f(x) |^2,dx \
    &=frac 1 2 int_mathbb{R} fracpi 2e^{-2|t|},dt end{align}
    The latter integral is elementary
    $$frac 1 2 int_mathbb{R} fracpi 2e^{-2|t|},dt = int_0^inftyfracpi 2e^{-2t},dt = frac pi 4$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      About your comment about the bounds, thanks. I thought because im trying to find the integral from $0$ to $infty$ then i will only look at this ray. So i need to write the $-infty,infty$ bounds everywhere and just devide by 2 because the function is even.
      $endgroup$
      – user123
      Dec 3 '18 at 5:53










    • $begingroup$
      where you wrote "Let us now use Plancherel to conclude i think the second equation it is suppose to be $overline{f}$ (f with hat, not sure what is the code for it.)
      $endgroup$
      – user123
      Dec 3 '18 at 5:56










    • $begingroup$
      @Liad Yes, I made a small correction, thanks! I removed the comment about the bounds since you were aware of that.
      $endgroup$
      – Shashi
      Dec 3 '18 at 6:55










    • $begingroup$
      alright, thanks for the answer, we indeed calculated the fourier transform of $e^{-alpha |x|}$ , i should've seen the connection. anyways, thanks a lot!
      $endgroup$
      – user123
      Dec 3 '18 at 6:58










    • $begingroup$
      @Liad you are welcome! I don't have time to look at it at the moment, but I will if I have time later in the day (if I won't forget it)
      $endgroup$
      – Shashi
      Dec 3 '18 at 7:06














    2












    2








    2





    $begingroup$

    Let me define $f$ as you have defined, namely $f(x) =(1+x^2)^{-1}$. In this case Plancherel doesn't make the evaluation much easier. However, in these kind of exercises they usually assume that you know that
    $$hat f(t) = frac{1}{sqrt{2pi} }int_mathbb{R} e^{-itx} f(x) , dx =sqrt{fracpi 2}e^{-|t|}. $$
    For example, finding the Fourier transform of that particular $f$ is asked in a previous exercise or it is given as an example in your notes. Anyways you can find a proof in this link.
    Let us now use Plancherel to conclude
    begin{align}
    int_0^infty frac 1 {(1+x^2)^2},dx&=frac 1 2 int_mathbb{R} |f(x) |^2,dx \
    &=frac 1 2 int_mathbb{R} |hat f(x) |^2,dx \
    &=frac 1 2 int_mathbb{R} fracpi 2e^{-2|t|},dt end{align}
    The latter integral is elementary
    $$frac 1 2 int_mathbb{R} fracpi 2e^{-2|t|},dt = int_0^inftyfracpi 2e^{-2t},dt = frac pi 4$$






    share|cite|improve this answer











    $endgroup$



    Let me define $f$ as you have defined, namely $f(x) =(1+x^2)^{-1}$. In this case Plancherel doesn't make the evaluation much easier. However, in these kind of exercises they usually assume that you know that
    $$hat f(t) = frac{1}{sqrt{2pi} }int_mathbb{R} e^{-itx} f(x) , dx =sqrt{fracpi 2}e^{-|t|}. $$
    For example, finding the Fourier transform of that particular $f$ is asked in a previous exercise or it is given as an example in your notes. Anyways you can find a proof in this link.
    Let us now use Plancherel to conclude
    begin{align}
    int_0^infty frac 1 {(1+x^2)^2},dx&=frac 1 2 int_mathbb{R} |f(x) |^2,dx \
    &=frac 1 2 int_mathbb{R} |hat f(x) |^2,dx \
    &=frac 1 2 int_mathbb{R} fracpi 2e^{-2|t|},dt end{align}
    The latter integral is elementary
    $$frac 1 2 int_mathbb{R} fracpi 2e^{-2|t|},dt = int_0^inftyfracpi 2e^{-2t},dt = frac pi 4$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 3 '18 at 6:54

























    answered Dec 2 '18 at 22:56









    ShashiShashi

    7,1731528




    7,1731528












    • $begingroup$
      About your comment about the bounds, thanks. I thought because im trying to find the integral from $0$ to $infty$ then i will only look at this ray. So i need to write the $-infty,infty$ bounds everywhere and just devide by 2 because the function is even.
      $endgroup$
      – user123
      Dec 3 '18 at 5:53










    • $begingroup$
      where you wrote "Let us now use Plancherel to conclude i think the second equation it is suppose to be $overline{f}$ (f with hat, not sure what is the code for it.)
      $endgroup$
      – user123
      Dec 3 '18 at 5:56










    • $begingroup$
      @Liad Yes, I made a small correction, thanks! I removed the comment about the bounds since you were aware of that.
      $endgroup$
      – Shashi
      Dec 3 '18 at 6:55










    • $begingroup$
      alright, thanks for the answer, we indeed calculated the fourier transform of $e^{-alpha |x|}$ , i should've seen the connection. anyways, thanks a lot!
      $endgroup$
      – user123
      Dec 3 '18 at 6:58










    • $begingroup$
      @Liad you are welcome! I don't have time to look at it at the moment, but I will if I have time later in the day (if I won't forget it)
      $endgroup$
      – Shashi
      Dec 3 '18 at 7:06


















    • $begingroup$
      About your comment about the bounds, thanks. I thought because im trying to find the integral from $0$ to $infty$ then i will only look at this ray. So i need to write the $-infty,infty$ bounds everywhere and just devide by 2 because the function is even.
      $endgroup$
      – user123
      Dec 3 '18 at 5:53










    • $begingroup$
      where you wrote "Let us now use Plancherel to conclude i think the second equation it is suppose to be $overline{f}$ (f with hat, not sure what is the code for it.)
      $endgroup$
      – user123
      Dec 3 '18 at 5:56










    • $begingroup$
      @Liad Yes, I made a small correction, thanks! I removed the comment about the bounds since you were aware of that.
      $endgroup$
      – Shashi
      Dec 3 '18 at 6:55










    • $begingroup$
      alright, thanks for the answer, we indeed calculated the fourier transform of $e^{-alpha |x|}$ , i should've seen the connection. anyways, thanks a lot!
      $endgroup$
      – user123
      Dec 3 '18 at 6:58










    • $begingroup$
      @Liad you are welcome! I don't have time to look at it at the moment, but I will if I have time later in the day (if I won't forget it)
      $endgroup$
      – Shashi
      Dec 3 '18 at 7:06
















    $begingroup$
    About your comment about the bounds, thanks. I thought because im trying to find the integral from $0$ to $infty$ then i will only look at this ray. So i need to write the $-infty,infty$ bounds everywhere and just devide by 2 because the function is even.
    $endgroup$
    – user123
    Dec 3 '18 at 5:53




    $begingroup$
    About your comment about the bounds, thanks. I thought because im trying to find the integral from $0$ to $infty$ then i will only look at this ray. So i need to write the $-infty,infty$ bounds everywhere and just devide by 2 because the function is even.
    $endgroup$
    – user123
    Dec 3 '18 at 5:53












    $begingroup$
    where you wrote "Let us now use Plancherel to conclude i think the second equation it is suppose to be $overline{f}$ (f with hat, not sure what is the code for it.)
    $endgroup$
    – user123
    Dec 3 '18 at 5:56




    $begingroup$
    where you wrote "Let us now use Plancherel to conclude i think the second equation it is suppose to be $overline{f}$ (f with hat, not sure what is the code for it.)
    $endgroup$
    – user123
    Dec 3 '18 at 5:56












    $begingroup$
    @Liad Yes, I made a small correction, thanks! I removed the comment about the bounds since you were aware of that.
    $endgroup$
    – Shashi
    Dec 3 '18 at 6:55




    $begingroup$
    @Liad Yes, I made a small correction, thanks! I removed the comment about the bounds since you were aware of that.
    $endgroup$
    – Shashi
    Dec 3 '18 at 6:55












    $begingroup$
    alright, thanks for the answer, we indeed calculated the fourier transform of $e^{-alpha |x|}$ , i should've seen the connection. anyways, thanks a lot!
    $endgroup$
    – user123
    Dec 3 '18 at 6:58




    $begingroup$
    alright, thanks for the answer, we indeed calculated the fourier transform of $e^{-alpha |x|}$ , i should've seen the connection. anyways, thanks a lot!
    $endgroup$
    – user123
    Dec 3 '18 at 6:58












    $begingroup$
    @Liad you are welcome! I don't have time to look at it at the moment, but I will if I have time later in the day (if I won't forget it)
    $endgroup$
    – Shashi
    Dec 3 '18 at 7:06




    $begingroup$
    @Liad you are welcome! I don't have time to look at it at the moment, but I will if I have time later in the day (if I won't forget it)
    $endgroup$
    – Shashi
    Dec 3 '18 at 7:06











    2












    $begingroup$

    This isn't the way that you asked for the integral to be computed, but it's a way that it could be computed.



    Consider the indefinite integral
    $$I_n=intfrac{mathrm{d}x}{(ax^2+b)^n}$$
    Integrating by parts with $mathrm{d}v=mathrm{d}x$ gives
    $$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{ax^2 mathrm{d}x}{(ax^2+b)^{n+1}}$$
    $$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{ax^2+b}{(ax^2+b)^{n+1}}mathrm{d}x-2bnintfrac{mathrm{d}x}{(ax^2+b)^{n+1}}$$
    $$I_n=frac{x}{(ax^2+b)^n}+2nI_n-2bnI_{n+1}$$
    $$2bnI_{n+1}=frac{x}{(ax^2+b)^n}+(2n-1)I_n$$
    $$I_{n+1}=frac{x}{2bn(ax^2+b)^n}+frac{2n-1}{2bn}I_n$$
    $$I_n=frac{x}{2b(n-1)(ax^2+b)^{n-1}}+frac{2n-3}{2b(n-1)}I_{n-1}$$
    Plugging in $a=1,,b=1,,n=2$, and evaluating form $x=0$ to $x=infty$ gives your integral as
    $$frac12int_0^infty frac{mathrm dx}{1+x^2}=fracpi4$$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      This isn't the way that you asked for the integral to be computed, but it's a way that it could be computed.



      Consider the indefinite integral
      $$I_n=intfrac{mathrm{d}x}{(ax^2+b)^n}$$
      Integrating by parts with $mathrm{d}v=mathrm{d}x$ gives
      $$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{ax^2 mathrm{d}x}{(ax^2+b)^{n+1}}$$
      $$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{ax^2+b}{(ax^2+b)^{n+1}}mathrm{d}x-2bnintfrac{mathrm{d}x}{(ax^2+b)^{n+1}}$$
      $$I_n=frac{x}{(ax^2+b)^n}+2nI_n-2bnI_{n+1}$$
      $$2bnI_{n+1}=frac{x}{(ax^2+b)^n}+(2n-1)I_n$$
      $$I_{n+1}=frac{x}{2bn(ax^2+b)^n}+frac{2n-1}{2bn}I_n$$
      $$I_n=frac{x}{2b(n-1)(ax^2+b)^{n-1}}+frac{2n-3}{2b(n-1)}I_{n-1}$$
      Plugging in $a=1,,b=1,,n=2$, and evaluating form $x=0$ to $x=infty$ gives your integral as
      $$frac12int_0^infty frac{mathrm dx}{1+x^2}=fracpi4$$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        This isn't the way that you asked for the integral to be computed, but it's a way that it could be computed.



        Consider the indefinite integral
        $$I_n=intfrac{mathrm{d}x}{(ax^2+b)^n}$$
        Integrating by parts with $mathrm{d}v=mathrm{d}x$ gives
        $$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{ax^2 mathrm{d}x}{(ax^2+b)^{n+1}}$$
        $$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{ax^2+b}{(ax^2+b)^{n+1}}mathrm{d}x-2bnintfrac{mathrm{d}x}{(ax^2+b)^{n+1}}$$
        $$I_n=frac{x}{(ax^2+b)^n}+2nI_n-2bnI_{n+1}$$
        $$2bnI_{n+1}=frac{x}{(ax^2+b)^n}+(2n-1)I_n$$
        $$I_{n+1}=frac{x}{2bn(ax^2+b)^n}+frac{2n-1}{2bn}I_n$$
        $$I_n=frac{x}{2b(n-1)(ax^2+b)^{n-1}}+frac{2n-3}{2b(n-1)}I_{n-1}$$
        Plugging in $a=1,,b=1,,n=2$, and evaluating form $x=0$ to $x=infty$ gives your integral as
        $$frac12int_0^infty frac{mathrm dx}{1+x^2}=fracpi4$$






        share|cite|improve this answer









        $endgroup$



        This isn't the way that you asked for the integral to be computed, but it's a way that it could be computed.



        Consider the indefinite integral
        $$I_n=intfrac{mathrm{d}x}{(ax^2+b)^n}$$
        Integrating by parts with $mathrm{d}v=mathrm{d}x$ gives
        $$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{ax^2 mathrm{d}x}{(ax^2+b)^{n+1}}$$
        $$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{ax^2+b}{(ax^2+b)^{n+1}}mathrm{d}x-2bnintfrac{mathrm{d}x}{(ax^2+b)^{n+1}}$$
        $$I_n=frac{x}{(ax^2+b)^n}+2nI_n-2bnI_{n+1}$$
        $$2bnI_{n+1}=frac{x}{(ax^2+b)^n}+(2n-1)I_n$$
        $$I_{n+1}=frac{x}{2bn(ax^2+b)^n}+frac{2n-1}{2bn}I_n$$
        $$I_n=frac{x}{2b(n-1)(ax^2+b)^{n-1}}+frac{2n-3}{2b(n-1)}I_{n-1}$$
        Plugging in $a=1,,b=1,,n=2$, and evaluating form $x=0$ to $x=infty$ gives your integral as
        $$frac12int_0^infty frac{mathrm dx}{1+x^2}=fracpi4$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 22:13









        clathratusclathratus

        3,862333




        3,862333






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3023226%2fcomputing-int-0-infty-dfrac11x22dx-using-plancherel%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
            Read more

            Puebla de Zaragoza

            Image
            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
            Read more

            Musa

            Image
            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
            Read more