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How to solve integrals like $int frac{e^{pm i 2pi t/T}}{a^{2}+t^{2}} dt$
$begingroup$
How can one solve integrals of the form
$$int frac{e^{pm i 2pi t/T}}{a^{2}+t^{2}} dt, $$
in general (indefinite) and in the definite cases with integration domains over the period ($T$) or infinity,
$$int_{-T/2}^{T/2}cdots text{or} int_{-infty}^{infty}cdots, $$
where $i=sqrt{-1}$ and $aneq0$ ?
Note: please note that I am looking for direct methods, not methods that rely on Fourier analysis and the Fourier duality property.
calculus integration
asked Jan 6 at 3:14
user135626user135626
432413
$endgroup$
add a comment |
$begingroup$
How can one solve integrals of the form
$$int frac{e^{pm i 2pi t/T}}{a^{2}+t^{2}} dt, $$
in general (indefinite) and in the definite cases with integration domains over the period ($T$) or infinity,
$$int_{-T/2}^{T/2}cdots text{or} int_{-infty}^{infty}cdots, $$
where $i=sqrt{-1}$ and $aneq0$ ?
Note: please note that I am looking for direct methods, not methods that rely on Fourier analysis and the Fourier duality property.
calculus integration
asked Jan 6 at 3:14
user135626user135626
432413
$endgroup$
$begingroup$
From what I recall from one of my classes last semester, I recall seeing Fourier transforms used to solve integrals like this, perhaps combined with the properties of the transform under convolution. This is just a shot in the dark, though, but it might be worth a shot. (EDIT: Well, with that edit forbidding Fourier stuff, I've got nothing.)
$endgroup$
– Eevee Trainer
Jan 6 at 3:16
$begingroup$
@Eevee I just added the note on Fourier analysis (you were quicker than my edit!). I am looking for more direct methods :) This would be more helpful to solve cases with different integration limits than infinity, and also to see an alternative way of reaching it.
$endgroup$
– user135626
Jan 6 at 3:17
$begingroup$
Are those limits meant to be $int_{-T/2}^{T/2}$ and $int_{-infty}^infty$?
$endgroup$
– John Doe
Jan 6 at 3:25
$begingroup$
This looks like something that can be solved through the use of contour integrals.
$endgroup$
– Aldon
Jan 6 at 3:26
$begingroup$
@JohnDoe Yes, corrected it, thanks.
$endgroup$
– user135626
Jan 6 at 4:28
add a comment |
$begingroup$
How can one solve integrals of the form
$$int frac{e^{pm i 2pi t/T}}{a^{2}+t^{2}} dt, $$
in general (indefinite) and in the definite cases with integration domains over the period ($T$) or infinity,
$$int_{-T/2}^{T/2}cdots text{or} int_{-infty}^{infty}cdots, $$
where $i=sqrt{-1}$ and $aneq0$ ?
Note: please note that I am looking for direct methods, not methods that rely on Fourier analysis and the Fourier duality property.
calculus integration
asked Jan 6 at 3:14
user135626user135626
432413
$endgroup$
How can one solve integrals of the form
$$int frac{e^{pm i 2pi t/T}}{a^{2}+t^{2}} dt, $$
in general (indefinite) and in the definite cases with integration domains over the period ($T$) or infinity,
$$int_{-T/2}^{T/2}cdots text{or} int_{-infty}^{infty}cdots, $$
where $i=sqrt{-1}$ and $aneq0$ ?
Note: please note that I am looking for direct methods, not methods that rely on Fourier analysis and the Fourier duality property.
calculus integration
calculus integration
asked Jan 6 at 3:14
user135626user135626
432413
asked Jan 6 at 3:14
user135626user135626
432413
edited Jan 6 at 4:27
user135626
asked Jan 6 at 3:14
user135626user135626
432413
asked Jan 6 at 3:14
user135626user135626
432413
asked Jan 6 at 3:14
user135626user135626
432413
432413
$begingroup$
From what I recall from one of my classes last semester, I recall seeing Fourier transforms used to solve integrals like this, perhaps combined with the properties of the transform under convolution. This is just a shot in the dark, though, but it might be worth a shot. (EDIT: Well, with that edit forbidding Fourier stuff, I've got nothing.)
$endgroup$
– Eevee Trainer
Jan 6 at 3:16
$begingroup$
@Eevee I just added the note on Fourier analysis (you were quicker than my edit!). I am looking for more direct methods :) This would be more helpful to solve cases with different integration limits than infinity, and also to see an alternative way of reaching it.
$endgroup$
– user135626
Jan 6 at 3:17
$begingroup$
Are those limits meant to be $int_{-T/2}^{T/2}$ and $int_{-infty}^infty$?
$endgroup$
– John Doe
Jan 6 at 3:25
$begingroup$
This looks like something that can be solved through the use of contour integrals.
$endgroup$
– Aldon
Jan 6 at 3:26
$begingroup$
@JohnDoe Yes, corrected it, thanks.
$endgroup$
– user135626
Jan 6 at 4:28
add a comment |
$begingroup$
From what I recall from one of my classes last semester, I recall seeing Fourier transforms used to solve integrals like this, perhaps combined with the properties of the transform under convolution. This is just a shot in the dark, though, but it might be worth a shot. (EDIT: Well, with that edit forbidding Fourier stuff, I've got nothing.)
$endgroup$
– Eevee Trainer
Jan 6 at 3:16
$begingroup$
@Eevee I just added the note on Fourier analysis (you were quicker than my edit!). I am looking for more direct methods :) This would be more helpful to solve cases with different integration limits than infinity, and also to see an alternative way of reaching it.
$endgroup$
– user135626
Jan 6 at 3:17
$begingroup$
Are those limits meant to be $int_{-T/2}^{T/2}$ and $int_{-infty}^infty$?
$endgroup$
– John Doe
Jan 6 at 3:25
$begingroup$
This looks like something that can be solved through the use of contour integrals.
$endgroup$
– Aldon
Jan 6 at 3:26
$begingroup$
@JohnDoe Yes, corrected it, thanks.
$endgroup$
– user135626
Jan 6 at 4:28
$begingroup$
From what I recall from one of my classes last semester, I recall seeing Fourier transforms used to solve integrals like this, perhaps combined with the properties of the transform under convolution. This is just a shot in the dark, though, but it might be worth a shot. (EDIT: Well, with that edit forbidding Fourier stuff, I've got nothing.)
$endgroup$
– Eevee Trainer
Jan 6 at 3:16
$begingroup$
From what I recall from one of my classes last semester, I recall seeing Fourier transforms used to solve integrals like this, perhaps combined with the properties of the transform under convolution. This is just a shot in the dark, though, but it might be worth a shot. (EDIT: Well, with that edit forbidding Fourier stuff, I've got nothing.)
$endgroup$
– Eevee Trainer
Jan 6 at 3:16
$begingroup$
@Eevee I just added the note on Fourier analysis (you were quicker than my edit!). I am looking for more direct methods :) This would be more helpful to solve cases with different integration limits than infinity, and also to see an alternative way of reaching it.
$endgroup$
– user135626
Jan 6 at 3:17
$begingroup$
@Eevee I just added the note on Fourier analysis (you were quicker than my edit!). I am looking for more direct methods :) This would be more helpful to solve cases with different integration limits than infinity, and also to see an alternative way of reaching it.
$endgroup$
– user135626
Jan 6 at 3:17
$begingroup$
Are those limits meant to be $int_{-T/2}^{T/2}$ and $int_{-infty}^infty$?
$endgroup$
– John Doe
Jan 6 at 3:25
$begingroup$
Are those limits meant to be $int_{-T/2}^{T/2}$ and $int_{-infty}^infty$?
$endgroup$
– John Doe
Jan 6 at 3:25
$begingroup$
This looks like something that can be solved through the use of contour integrals.
$endgroup$
– Aldon
Jan 6 at 3:26
$begingroup$
This looks like something that can be solved through the use of contour integrals.
$endgroup$
– Aldon
Jan 6 at 3:26
$begingroup$
@JohnDoe Yes, corrected it, thanks.
$endgroup$
– user135626
Jan 6 at 4:28
$begingroup$
@JohnDoe Yes, corrected it, thanks.
$endgroup$
– user135626
Jan 6 at 4:28
add a comment |
2 Answers
2
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oldest
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$begingroup$
For the definite integral case, if we have $$int_{-infty}^inftyfrac{e^{2pi i t/T}}{a^2+t^2}mathrm dt$$then we can evaluate this using contour integration by picking a semi-circular contour in the upper half plane. This encloses a simple pole at $t=ai$, and it can be shown that as the radius of the semi-circle tends to $infty$, the contribution to the integral from the curved part of the contour is zero. So the integral is equal to $$2pi i,text{Res}left(frac{e^{2pi i t/T}}{a^2+t^2},airight)=2pi i frac{e^{-2pi a/T}}{2ai}=fracpi ae^{-2pi a/T}$$
For $$int_{-T/2}^{T/2}frac{e^{2pi i t/T}}{a^2+t^2}mathrm dt$$we have that this integral equals $$int_0^pifrac{e^{pi i e^{itheta}}}{a^2+T^2/4cdot e^{2itheta}}(T/2) e^{itheta}imathrm dtheta+fracpi ae^{-2pi a/T}textbf 1_{[0,T/2]}(|a|)$$ where $textbf 1_{A}(x)=begin{cases}1&xin A\0&xnotin Aend{cases}$ is an indicator function. I am not sure of a simple way to compute this integral.
The indefinite case is, of course, the hardest.
answered Jan 6 at 3:30
John DoeJohn Doe
11.1k11238
$endgroup$
add a comment |
$begingroup$
The indefinite integral can't be done using elementary functions. It can be expressed in terms of special functions. Thus Maple writes it as
$$ {frac {i}{2a} left( {{rm e}^{-{frac {2pi,a}{T}}}}{rm Ei}_1
left( -{frac {2pi, left( it+a right) }{T}} right) -{
{rm e}^{{frac {2pi,a}{T}}}}{rm Ei}_1 left(-{frac {2pi,
left( it-a right) }{T}} right) right) }$$
EDIT: The fact that there is no elementary antiderivative can be proven. It follows from the Risch theory (actually this case is a theorem of Liouville, IIRC) that the only cases where $int r(t) exp(a t); dt$ is elementary, where $a$ is a nonzero constant and $r(t)$ is a rational function, is where the antiderivatives are of the form $R(t) exp(a t) + c$, and then $r(t) = R'(t) + a R(t)$. In particular, this can't happen if $r(t)$ has a simple pole.
answered Jan 6 at 4:12
Robert IsraelRobert Israel
321k23210462
$endgroup$
add a comment |
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2 Answers
2
active
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2 Answers
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$begingroup$
For the definite integral case, if we have $$int_{-infty}^inftyfrac{e^{2pi i t/T}}{a^2+t^2}mathrm dt$$then we can evaluate this using contour integration by picking a semi-circular contour in the upper half plane. This encloses a simple pole at $t=ai$, and it can be shown that as the radius of the semi-circle tends to $infty$, the contribution to the integral from the curved part of the contour is zero. So the integral is equal to $$2pi i,text{Res}left(frac{e^{2pi i t/T}}{a^2+t^2},airight)=2pi i frac{e^{-2pi a/T}}{2ai}=fracpi ae^{-2pi a/T}$$
For $$int_{-T/2}^{T/2}frac{e^{2pi i t/T}}{a^2+t^2}mathrm dt$$we have that this integral equals $$int_0^pifrac{e^{pi i e^{itheta}}}{a^2+T^2/4cdot e^{2itheta}}(T/2) e^{itheta}imathrm dtheta+fracpi ae^{-2pi a/T}textbf 1_{[0,T/2]}(|a|)$$ where $textbf 1_{A}(x)=begin{cases}1&xin A\0&xnotin Aend{cases}$ is an indicator function. I am not sure of a simple way to compute this integral.
The indefinite case is, of course, the hardest.
answered Jan 6 at 3:30
John DoeJohn Doe
11.1k11238
$endgroup$
add a comment |
$begingroup$
For the definite integral case, if we have $$int_{-infty}^inftyfrac{e^{2pi i t/T}}{a^2+t^2}mathrm dt$$then we can evaluate this using contour integration by picking a semi-circular contour in the upper half plane. This encloses a simple pole at $t=ai$, and it can be shown that as the radius of the semi-circle tends to $infty$, the contribution to the integral from the curved part of the contour is zero. So the integral is equal to $$2pi i,text{Res}left(frac{e^{2pi i t/T}}{a^2+t^2},airight)=2pi i frac{e^{-2pi a/T}}{2ai}=fracpi ae^{-2pi a/T}$$
For $$int_{-T/2}^{T/2}frac{e^{2pi i t/T}}{a^2+t^2}mathrm dt$$we have that this integral equals $$int_0^pifrac{e^{pi i e^{itheta}}}{a^2+T^2/4cdot e^{2itheta}}(T/2) e^{itheta}imathrm dtheta+fracpi ae^{-2pi a/T}textbf 1_{[0,T/2]}(|a|)$$ where $textbf 1_{A}(x)=begin{cases}1&xin A\0&xnotin Aend{cases}$ is an indicator function. I am not sure of a simple way to compute this integral.
The indefinite case is, of course, the hardest.
answered Jan 6 at 3:30
John DoeJohn Doe
11.1k11238
$endgroup$
add a comment |
$begingroup$
For the definite integral case, if we have $$int_{-infty}^inftyfrac{e^{2pi i t/T}}{a^2+t^2}mathrm dt$$then we can evaluate this using contour integration by picking a semi-circular contour in the upper half plane. This encloses a simple pole at $t=ai$, and it can be shown that as the radius of the semi-circle tends to $infty$, the contribution to the integral from the curved part of the contour is zero. So the integral is equal to $$2pi i,text{Res}left(frac{e^{2pi i t/T}}{a^2+t^2},airight)=2pi i frac{e^{-2pi a/T}}{2ai}=fracpi ae^{-2pi a/T}$$
For $$int_{-T/2}^{T/2}frac{e^{2pi i t/T}}{a^2+t^2}mathrm dt$$we have that this integral equals $$int_0^pifrac{e^{pi i e^{itheta}}}{a^2+T^2/4cdot e^{2itheta}}(T/2) e^{itheta}imathrm dtheta+fracpi ae^{-2pi a/T}textbf 1_{[0,T/2]}(|a|)$$ where $textbf 1_{A}(x)=begin{cases}1&xin A\0&xnotin Aend{cases}$ is an indicator function. I am not sure of a simple way to compute this integral.
The indefinite case is, of course, the hardest.
answered Jan 6 at 3:30
John DoeJohn Doe
11.1k11238
$endgroup$
For the definite integral case, if we have $$int_{-infty}^inftyfrac{e^{2pi i t/T}}{a^2+t^2}mathrm dt$$then we can evaluate this using contour integration by picking a semi-circular contour in the upper half plane. This encloses a simple pole at $t=ai$, and it can be shown that as the radius of the semi-circle tends to $infty$, the contribution to the integral from the curved part of the contour is zero. So the integral is equal to $$2pi i,text{Res}left(frac{e^{2pi i t/T}}{a^2+t^2},airight)=2pi i frac{e^{-2pi a/T}}{2ai}=fracpi ae^{-2pi a/T}$$
For $$int_{-T/2}^{T/2}frac{e^{2pi i t/T}}{a^2+t^2}mathrm dt$$we have that this integral equals $$int_0^pifrac{e^{pi i e^{itheta}}}{a^2+T^2/4cdot e^{2itheta}}(T/2) e^{itheta}imathrm dtheta+fracpi ae^{-2pi a/T}textbf 1_{[0,T/2]}(|a|)$$ where $textbf 1_{A}(x)=begin{cases}1&xin A\0&xnotin Aend{cases}$ is an indicator function. I am not sure of a simple way to compute this integral.
The indefinite case is, of course, the hardest.
answered Jan 6 at 3:30
John DoeJohn Doe
11.1k11238
edited Jan 6 at 3:47
answered Jan 6 at 3:30
John DoeJohn Doe
11.1k11238
answered Jan 6 at 3:30
John DoeJohn Doe
11.1k11238
answered Jan 6 at 3:30
John DoeJohn Doe
11.1k11238
11.1k11238
add a comment |
add a comment |
$begingroup$
The indefinite integral can't be done using elementary functions. It can be expressed in terms of special functions. Thus Maple writes it as
$$ {frac {i}{2a} left( {{rm e}^{-{frac {2pi,a}{T}}}}{rm Ei}_1
left( -{frac {2pi, left( it+a right) }{T}} right) -{
{rm e}^{{frac {2pi,a}{T}}}}{rm Ei}_1 left(-{frac {2pi,
left( it-a right) }{T}} right) right) }$$
EDIT: The fact that there is no elementary antiderivative can be proven. It follows from the Risch theory (actually this case is a theorem of Liouville, IIRC) that the only cases where $int r(t) exp(a t); dt$ is elementary, where $a$ is a nonzero constant and $r(t)$ is a rational function, is where the antiderivatives are of the form $R(t) exp(a t) + c$, and then $r(t) = R'(t) + a R(t)$. In particular, this can't happen if $r(t)$ has a simple pole.
answered Jan 6 at 4:12
Robert IsraelRobert Israel
321k23210462
$endgroup$
add a comment |
$begingroup$
The indefinite integral can't be done using elementary functions. It can be expressed in terms of special functions. Thus Maple writes it as
$$ {frac {i}{2a} left( {{rm e}^{-{frac {2pi,a}{T}}}}{rm Ei}_1
left( -{frac {2pi, left( it+a right) }{T}} right) -{
{rm e}^{{frac {2pi,a}{T}}}}{rm Ei}_1 left(-{frac {2pi,
left( it-a right) }{T}} right) right) }$$
EDIT: The fact that there is no elementary antiderivative can be proven. It follows from the Risch theory (actually this case is a theorem of Liouville, IIRC) that the only cases where $int r(t) exp(a t); dt$ is elementary, where $a$ is a nonzero constant and $r(t)$ is a rational function, is where the antiderivatives are of the form $R(t) exp(a t) + c$, and then $r(t) = R'(t) + a R(t)$. In particular, this can't happen if $r(t)$ has a simple pole.
answered Jan 6 at 4:12
Robert IsraelRobert Israel
321k23210462
$endgroup$
add a comment |
$begingroup$
The indefinite integral can't be done using elementary functions. It can be expressed in terms of special functions. Thus Maple writes it as
$$ {frac {i}{2a} left( {{rm e}^{-{frac {2pi,a}{T}}}}{rm Ei}_1
left( -{frac {2pi, left( it+a right) }{T}} right) -{
{rm e}^{{frac {2pi,a}{T}}}}{rm Ei}_1 left(-{frac {2pi,
left( it-a right) }{T}} right) right) }$$
EDIT: The fact that there is no elementary antiderivative can be proven. It follows from the Risch theory (actually this case is a theorem of Liouville, IIRC) that the only cases where $int r(t) exp(a t); dt$ is elementary, where $a$ is a nonzero constant and $r(t)$ is a rational function, is where the antiderivatives are of the form $R(t) exp(a t) + c$, and then $r(t) = R'(t) + a R(t)$. In particular, this can't happen if $r(t)$ has a simple pole.
answered Jan 6 at 4:12
Robert IsraelRobert Israel
321k23210462
$endgroup$
The indefinite integral can't be done using elementary functions. It can be expressed in terms of special functions. Thus Maple writes it as
$$ {frac {i}{2a} left( {{rm e}^{-{frac {2pi,a}{T}}}}{rm Ei}_1
left( -{frac {2pi, left( it+a right) }{T}} right) -{
{rm e}^{{frac {2pi,a}{T}}}}{rm Ei}_1 left(-{frac {2pi,
left( it-a right) }{T}} right) right) }$$
EDIT: The fact that there is no elementary antiderivative can be proven. It follows from the Risch theory (actually this case is a theorem of Liouville, IIRC) that the only cases where $int r(t) exp(a t); dt$ is elementary, where $a$ is a nonzero constant and $r(t)$ is a rational function, is where the antiderivatives are of the form $R(t) exp(a t) + c$, and then $r(t) = R'(t) + a R(t)$. In particular, this can't happen if $r(t)$ has a simple pole.
answered Jan 6 at 4:12
Robert IsraelRobert Israel
321k23210462
edited Jan 6 at 15:41
answered Jan 6 at 4:12
Robert IsraelRobert Israel
321k23210462
answered Jan 6 at 4:12
Robert IsraelRobert Israel
321k23210462
answered Jan 6 at 4:12
Robert IsraelRobert Israel
321k23210462
321k23210462
add a comment |
add a comment |
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$begingroup$
From what I recall from one of my classes last semester, I recall seeing Fourier transforms used to solve integrals like this, perhaps combined with the properties of the transform under convolution. This is just a shot in the dark, though, but it might be worth a shot. (EDIT: Well, with that edit forbidding Fourier stuff, I've got nothing.)
$endgroup$
– Eevee Trainer
Jan 6 at 3:16
$begingroup$
@Eevee I just added the note on Fourier analysis (you were quicker than my edit!). I am looking for more direct methods :) This would be more helpful to solve cases with different integration limits than infinity, and also to see an alternative way of reaching it.
$endgroup$
– user135626
Jan 6 at 3:17
$begingroup$
Are those limits meant to be $int_{-T/2}^{T/2}$ and $int_{-infty}^infty$?
$endgroup$
– John Doe
Jan 6 at 3:25
$begingroup$
This looks like something that can be solved through the use of contour integrals.
$endgroup$
– Aldon
Jan 6 at 3:26
$begingroup$
@JohnDoe Yes, corrected it, thanks.
$endgroup$
– user135626
Jan 6 at 4:28