Show that the following integral is zero.
$begingroup$
This problem is very challenging. The integral cannot be be solved in terms of elementary functions.
Show the following integral vanishes:
begin{equation}
displaystyle int_{-3}^{-1} frac{1}{e^{0.5x+e^{2-x}}+e^{0.5x}}f(x)dx+displaystyle int_{1}^{3} frac{1}{e^{0.5x+e^{2-x}}+e^{0.5x}}f(x) dx
end{equation}
where
begin{equation}
f(x)=displaystyle int_{-3}^{-1} e^{frac{1}{t^2+4t+3}}cos{2pi x t} dt+ displaystyle int_{1}^{3} e^{frac{1}{t^2-4t+3}}cos{2pi x t} dt
end{equation}
Perhaps, one of the ways would be to find the series of $f(x)$ and plug in the given integral, but I still find challenges there?
real-analysis calculus definite-integrals
asked Dec 22 '18 at 0:20
ershersh
573113
$endgroup$
|
show 4 more comments
$begingroup$
This problem is very challenging. The integral cannot be be solved in terms of elementary functions.
Show the following integral vanishes:
begin{equation}
displaystyle int_{-3}^{-1} frac{1}{e^{0.5x+e^{2-x}}+e^{0.5x}}f(x)dx+displaystyle int_{1}^{3} frac{1}{e^{0.5x+e^{2-x}}+e^{0.5x}}f(x) dx
end{equation}
where
begin{equation}
f(x)=displaystyle int_{-3}^{-1} e^{frac{1}{t^2+4t+3}}cos{2pi x t} dt+ displaystyle int_{1}^{3} e^{frac{1}{t^2-4t+3}}cos{2pi x t} dt
end{equation}
Perhaps, one of the ways would be to find the series of $f(x)$ and plug in the given integral, but I still find challenges there?
real-analysis calculus definite-integrals
asked Dec 22 '18 at 0:20
ershersh
573113
$endgroup$
$begingroup$
Have you checked this numerically?
$endgroup$
– ablmf
Dec 22 '18 at 1:18
$begingroup$
YES, The answer is close to zero, in Sage and in Mathematica. We can't reduce the error much. I also have a long proof that this is true, but that is not the direct substitution.
$endgroup$
– ersh
Dec 22 '18 at 1:30
$begingroup$
Can't you simplify $f(x)$ a bit with the substitution $u=-t$? Not sure that is going to help you in the long run, but it should just be twice one of the two integrals.
$endgroup$
– DanielOfJack
Dec 22 '18 at 1:51
$begingroup$
Also, if you have a proof that this is true, what precisely are you asking for?
$endgroup$
– DanielOfJack
Dec 22 '18 at 1:54
$begingroup$
@DanielOfJack The proof I have is indirect in the sense that it uses many theorems, implying the integral vanishes for this $f(x)$.
$endgroup$
– ersh
Dec 22 '18 at 3:30
|
show 4 more comments
$begingroup$
This problem is very challenging. The integral cannot be be solved in terms of elementary functions.
Show the following integral vanishes:
begin{equation}
displaystyle int_{-3}^{-1} frac{1}{e^{0.5x+e^{2-x}}+e^{0.5x}}f(x)dx+displaystyle int_{1}^{3} frac{1}{e^{0.5x+e^{2-x}}+e^{0.5x}}f(x) dx
end{equation}
where
begin{equation}
f(x)=displaystyle int_{-3}^{-1} e^{frac{1}{t^2+4t+3}}cos{2pi x t} dt+ displaystyle int_{1}^{3} e^{frac{1}{t^2-4t+3}}cos{2pi x t} dt
end{equation}
Perhaps, one of the ways would be to find the series of $f(x)$ and plug in the given integral, but I still find challenges there?
real-analysis calculus definite-integrals
asked Dec 22 '18 at 0:20
ershersh
573113
$endgroup$
This problem is very challenging. The integral cannot be be solved in terms of elementary functions.
Show the following integral vanishes:
begin{equation}
displaystyle int_{-3}^{-1} frac{1}{e^{0.5x+e^{2-x}}+e^{0.5x}}f(x)dx+displaystyle int_{1}^{3} frac{1}{e^{0.5x+e^{2-x}}+e^{0.5x}}f(x) dx
end{equation}
where
begin{equation}
f(x)=displaystyle int_{-3}^{-1} e^{frac{1}{t^2+4t+3}}cos{2pi x t} dt+ displaystyle int_{1}^{3} e^{frac{1}{t^2-4t+3}}cos{2pi x t} dt
end{equation}
Perhaps, one of the ways would be to find the series of $f(x)$ and plug in the given integral, but I still find challenges there?
real-analysis calculus definite-integrals
real-analysis calculus definite-integrals
asked Dec 22 '18 at 0:20
ershersh
573113
asked Dec 22 '18 at 0:20
ershersh
573113
asked Dec 22 '18 at 0:20
ershersh
573113
asked Dec 22 '18 at 0:20
ershersh
573113
asked Dec 22 '18 at 0:20
ershersh
573113
573113
$begingroup$
Have you checked this numerically?
$endgroup$
– ablmf
Dec 22 '18 at 1:18
$begingroup$
YES, The answer is close to zero, in Sage and in Mathematica. We can't reduce the error much. I also have a long proof that this is true, but that is not the direct substitution.
$endgroup$
– ersh
Dec 22 '18 at 1:30
$begingroup$
Can't you simplify $f(x)$ a bit with the substitution $u=-t$? Not sure that is going to help you in the long run, but it should just be twice one of the two integrals.
$endgroup$
– DanielOfJack
Dec 22 '18 at 1:51
$begingroup$
Also, if you have a proof that this is true, what precisely are you asking for?
$endgroup$
– DanielOfJack
Dec 22 '18 at 1:54
$begingroup$
@DanielOfJack The proof I have is indirect in the sense that it uses many theorems, implying the integral vanishes for this $f(x)$.
$endgroup$
– ersh
Dec 22 '18 at 3:30
|
show 4 more comments
$begingroup$
Have you checked this numerically?
$endgroup$
– ablmf
Dec 22 '18 at 1:18
$begingroup$
YES, The answer is close to zero, in Sage and in Mathematica. We can't reduce the error much. I also have a long proof that this is true, but that is not the direct substitution.
$endgroup$
– ersh
Dec 22 '18 at 1:30
$begingroup$
Can't you simplify $f(x)$ a bit with the substitution $u=-t$? Not sure that is going to help you in the long run, but it should just be twice one of the two integrals.
$endgroup$
– DanielOfJack
Dec 22 '18 at 1:51
$begingroup$
Also, if you have a proof that this is true, what precisely are you asking for?
$endgroup$
– DanielOfJack
Dec 22 '18 at 1:54
$begingroup$
@DanielOfJack The proof I have is indirect in the sense that it uses many theorems, implying the integral vanishes for this $f(x)$.
$endgroup$
– ersh
Dec 22 '18 at 3:30
$begingroup$
Have you checked this numerically?
$endgroup$
– ablmf
Dec 22 '18 at 1:18
$begingroup$
Have you checked this numerically?
$endgroup$
– ablmf
Dec 22 '18 at 1:18
$begingroup$
YES, The answer is close to zero, in Sage and in Mathematica. We can't reduce the error much. I also have a long proof that this is true, but that is not the direct substitution.
$endgroup$
– ersh
Dec 22 '18 at 1:30
$begingroup$
YES, The answer is close to zero, in Sage and in Mathematica. We can't reduce the error much. I also have a long proof that this is true, but that is not the direct substitution.
$endgroup$
– ersh
Dec 22 '18 at 1:30
$begingroup$
Can't you simplify $f(x)$ a bit with the substitution $u=-t$? Not sure that is going to help you in the long run, but it should just be twice one of the two integrals.
$endgroup$
– DanielOfJack
Dec 22 '18 at 1:51
$begingroup$
Can't you simplify $f(x)$ a bit with the substitution $u=-t$? Not sure that is going to help you in the long run, but it should just be twice one of the two integrals.
$endgroup$
– DanielOfJack
Dec 22 '18 at 1:51
$begingroup$
Also, if you have a proof that this is true, what precisely are you asking for?
$endgroup$
– DanielOfJack
Dec 22 '18 at 1:54
$begingroup$
Also, if you have a proof that this is true, what precisely are you asking for?
$endgroup$
– DanielOfJack
Dec 22 '18 at 1:54
$begingroup$
@DanielOfJack The proof I have is indirect in the sense that it uses many theorems, implying the integral vanishes for this $f(x)$.
$endgroup$
– ersh
Dec 22 '18 at 3:30
$begingroup$
@DanielOfJack The proof I have is indirect in the sense that it uses many theorems, implying the integral vanishes for this $f(x)$.
$endgroup$
– ersh
Dec 22 '18 at 3:30
|
show 4 more comments
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$begingroup$
Have you checked this numerically?
$endgroup$
– ablmf
Dec 22 '18 at 1:18
$begingroup$
YES, The answer is close to zero, in Sage and in Mathematica. We can't reduce the error much. I also have a long proof that this is true, but that is not the direct substitution.
$endgroup$
– ersh
Dec 22 '18 at 1:30
$begingroup$
Can't you simplify $f(x)$ a bit with the substitution $u=-t$? Not sure that is going to help you in the long run, but it should just be twice one of the two integrals.
$endgroup$
– DanielOfJack
Dec 22 '18 at 1:51
$begingroup$
Also, if you have a proof that this is true, what precisely are you asking for?
$endgroup$
– DanielOfJack
Dec 22 '18 at 1:54
$begingroup$
@DanielOfJack The proof I have is indirect in the sense that it uses many theorems, implying the integral vanishes for this $f(x)$.
$endgroup$
– ersh
Dec 22 '18 at 3:30