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









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    2 Answers
    2






    active

    oldest

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    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






























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