Mixing
Evaluating $int_{-infty}^{infty}text{sech}^2(x)cos(x),dx$
$begingroup$
I'm studying a paper where an integral of similar form to this one appears:
$$int_{-infty}^{infty}text{sech}^2(x)cos(x),dx$$
The authors only show the result, which involves a hyperbolic cosecant function with a $pi$ in its argument. So, I assume that they considered a closed path and used the residuals theorem. I know that the poles of the integrand function are $z=ileft(frac{pi}{2}+kpiright),,kinmathbb{Z}$. I tried to solve the integral by myself, but I don't know which path to choose. Do you have any hints to solve it?
improper-integrals
asked Jan 9 at 12:40
Élio PereiraÉlio Pereira
402514
$endgroup$
add a comment |
$begingroup$
I'm studying a paper where an integral of similar form to this one appears:
$$int_{-infty}^{infty}text{sech}^2(x)cos(x),dx$$
The authors only show the result, which involves a hyperbolic cosecant function with a $pi$ in its argument. So, I assume that they considered a closed path and used the residuals theorem. I know that the poles of the integrand function are $z=ileft(frac{pi}{2}+kpiright),,kinmathbb{Z}$. I tried to solve the integral by myself, but I don't know which path to choose. Do you have any hints to solve it?
improper-integrals
asked Jan 9 at 12:40
Élio PereiraÉlio Pereira
402514
$endgroup$
add a comment |
$begingroup$
I'm studying a paper where an integral of similar form to this one appears:
$$int_{-infty}^{infty}text{sech}^2(x)cos(x),dx$$
The authors only show the result, which involves a hyperbolic cosecant function with a $pi$ in its argument. So, I assume that they considered a closed path and used the residuals theorem. I know that the poles of the integrand function are $z=ileft(frac{pi}{2}+kpiright),,kinmathbb{Z}$. I tried to solve the integral by myself, but I don't know which path to choose. Do you have any hints to solve it?
improper-integrals
asked Jan 9 at 12:40
Élio PereiraÉlio Pereira
402514
$endgroup$
I'm studying a paper where an integral of similar form to this one appears:
$$int_{-infty}^{infty}text{sech}^2(x)cos(x),dx$$
The authors only show the result, which involves a hyperbolic cosecant function with a $pi$ in its argument. So, I assume that they considered a closed path and used the residuals theorem. I know that the poles of the integrand function are $z=ileft(frac{pi}{2}+kpiright),,kinmathbb{Z}$. I tried to solve the integral by myself, but I don't know which path to choose. Do you have any hints to solve it?
improper-integrals
improper-integrals
asked Jan 9 at 12:40
Élio PereiraÉlio Pereira
402514
asked Jan 9 at 12:40
Élio PereiraÉlio Pereira
402514
edited Jan 9 at 13:31
Élio Pereira
asked Jan 9 at 12:40
Élio PereiraÉlio Pereira
402514
asked Jan 9 at 12:40
Élio PereiraÉlio Pereira
402514
asked Jan 9 at 12:40
Élio PereiraÉlio Pereira
402514
402514
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
Use $operatorname{Re} e^{ix} =cos x$. Then choose a contour that encircles the upper half-plane.
By using the residue theorem, we get the result
$$I= int_{-infty}^{infty} frac{cos x}{cosh^2 x},dx = operatorname{Re} sum_{k=0}^infty 2pi i,operatorname{Res}_{x=ipi (k+1/2)} left(frac{e^{ix}}{cosh^2 x}right).$$
We can calculate the residues as (the poles are of second order and $cosh^{-2}(x) approx -(x-x_0)^{-2}$ close to the pole at $x_0$.)
$$operatorname{Res}_{x=ipi (k+1/2)} left(frac{e^{ix}}{cosh^2 x}right) = - frac{d}{dx} e^{ix} Big|_{x=ipi (k+1/2)} = -i e^{-pi(k+1/2)},.$$
The integral then assumes the form
$$ I= 2pi sum_{k=0}^infty e^{-pi(k+1/2)} = frac{2pi e^{-pi/2} }{1-e^{-pi}} = frac{pi}{sinh(pi/2)},.$$
answered Jan 9 at 12:44
FabianFabian
19.7k3674
$endgroup$
$begingroup$
Aren't the poles of second order? The hyperbolic cosine function is squared.
$endgroup$
– Élio Pereira
Jan 9 at 13:15
1
$begingroup$
@ÉlioPereira: thanks, I will correct it.
$endgroup$
– Fabian
Jan 9 at 13:24
$begingroup$
Thank you very much for your help.
$endgroup$
– Élio Pereira
Jan 9 at 13:32
$begingroup$
Can you help me with just one more thing? How could I show that the integral on the arc is null?
$endgroup$
– Élio Pereira
Jan 9 at 16:32
1
$begingroup$
@ÉlioPereira: there are different ways to show that. The most straightforward way might be that $|e^{i z}/cosh(z)^2| = 2e^{-y}/(cos 2 y + cosh 2x) leq 2 e^{-y}/(cosh2 x -1)$ for $z=x+i y$. On the arc, we have that $yto+infty$ fast enough that the integral vanishes. To show this technically, you might have to distinguish points $z=r e^{itheta}$ with $|theta -pi/2| < epsilon$ from the rest.
$endgroup$
– Fabian
Jan 9 at 17:56
|
show 5 more comments
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
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active
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votes
$begingroup$
Hint:
Use $operatorname{Re} e^{ix} =cos x$. Then choose a contour that encircles the upper half-plane.
By using the residue theorem, we get the result
$$I= int_{-infty}^{infty} frac{cos x}{cosh^2 x},dx = operatorname{Re} sum_{k=0}^infty 2pi i,operatorname{Res}_{x=ipi (k+1/2)} left(frac{e^{ix}}{cosh^2 x}right).$$
We can calculate the residues as (the poles are of second order and $cosh^{-2}(x) approx -(x-x_0)^{-2}$ close to the pole at $x_0$.)
$$operatorname{Res}_{x=ipi (k+1/2)} left(frac{e^{ix}}{cosh^2 x}right) = - frac{d}{dx} e^{ix} Big|_{x=ipi (k+1/2)} = -i e^{-pi(k+1/2)},.$$
The integral then assumes the form
$$ I= 2pi sum_{k=0}^infty e^{-pi(k+1/2)} = frac{2pi e^{-pi/2} }{1-e^{-pi}} = frac{pi}{sinh(pi/2)},.$$
answered Jan 9 at 12:44
FabianFabian
19.7k3674
$endgroup$
$begingroup$
Aren't the poles of second order? The hyperbolic cosine function is squared.
$endgroup$
– Élio Pereira
Jan 9 at 13:15
1
$begingroup$
@ÉlioPereira: thanks, I will correct it.
$endgroup$
– Fabian
Jan 9 at 13:24
$begingroup$
Thank you very much for your help.
$endgroup$
– Élio Pereira
Jan 9 at 13:32
$begingroup$
Can you help me with just one more thing? How could I show that the integral on the arc is null?
$endgroup$
– Élio Pereira
Jan 9 at 16:32
1
$begingroup$
@ÉlioPereira: there are different ways to show that. The most straightforward way might be that $|e^{i z}/cosh(z)^2| = 2e^{-y}/(cos 2 y + cosh 2x) leq 2 e^{-y}/(cosh2 x -1)$ for $z=x+i y$. On the arc, we have that $yto+infty$ fast enough that the integral vanishes. To show this technically, you might have to distinguish points $z=r e^{itheta}$ with $|theta -pi/2| < epsilon$ from the rest.
$endgroup$
– Fabian
Jan 9 at 17:56
|
show 5 more comments
$begingroup$
Hint:
Use $operatorname{Re} e^{ix} =cos x$. Then choose a contour that encircles the upper half-plane.
By using the residue theorem, we get the result
$$I= int_{-infty}^{infty} frac{cos x}{cosh^2 x},dx = operatorname{Re} sum_{k=0}^infty 2pi i,operatorname{Res}_{x=ipi (k+1/2)} left(frac{e^{ix}}{cosh^2 x}right).$$
We can calculate the residues as (the poles are of second order and $cosh^{-2}(x) approx -(x-x_0)^{-2}$ close to the pole at $x_0$.)
$$operatorname{Res}_{x=ipi (k+1/2)} left(frac{e^{ix}}{cosh^2 x}right) = - frac{d}{dx} e^{ix} Big|_{x=ipi (k+1/2)} = -i e^{-pi(k+1/2)},.$$
The integral then assumes the form
$$ I= 2pi sum_{k=0}^infty e^{-pi(k+1/2)} = frac{2pi e^{-pi/2} }{1-e^{-pi}} = frac{pi}{sinh(pi/2)},.$$
answered Jan 9 at 12:44
FabianFabian
19.7k3674
$endgroup$
$begingroup$
Aren't the poles of second order? The hyperbolic cosine function is squared.
$endgroup$
– Élio Pereira
Jan 9 at 13:15
1
$begingroup$
@ÉlioPereira: thanks, I will correct it.
$endgroup$
– Fabian
Jan 9 at 13:24
$begingroup$
Thank you very much for your help.
$endgroup$
– Élio Pereira
Jan 9 at 13:32
$begingroup$
Can you help me with just one more thing? How could I show that the integral on the arc is null?
$endgroup$
– Élio Pereira
Jan 9 at 16:32
1
$begingroup$
@ÉlioPereira: there are different ways to show that. The most straightforward way might be that $|e^{i z}/cosh(z)^2| = 2e^{-y}/(cos 2 y + cosh 2x) leq 2 e^{-y}/(cosh2 x -1)$ for $z=x+i y$. On the arc, we have that $yto+infty$ fast enough that the integral vanishes. To show this technically, you might have to distinguish points $z=r e^{itheta}$ with $|theta -pi/2| < epsilon$ from the rest.
$endgroup$
– Fabian
Jan 9 at 17:56
|
show 5 more comments
$begingroup$
Hint:
Use $operatorname{Re} e^{ix} =cos x$. Then choose a contour that encircles the upper half-plane.
By using the residue theorem, we get the result
$$I= int_{-infty}^{infty} frac{cos x}{cosh^2 x},dx = operatorname{Re} sum_{k=0}^infty 2pi i,operatorname{Res}_{x=ipi (k+1/2)} left(frac{e^{ix}}{cosh^2 x}right).$$
We can calculate the residues as (the poles are of second order and $cosh^{-2}(x) approx -(x-x_0)^{-2}$ close to the pole at $x_0$.)
$$operatorname{Res}_{x=ipi (k+1/2)} left(frac{e^{ix}}{cosh^2 x}right) = - frac{d}{dx} e^{ix} Big|_{x=ipi (k+1/2)} = -i e^{-pi(k+1/2)},.$$
The integral then assumes the form
$$ I= 2pi sum_{k=0}^infty e^{-pi(k+1/2)} = frac{2pi e^{-pi/2} }{1-e^{-pi}} = frac{pi}{sinh(pi/2)},.$$
answered Jan 9 at 12:44
FabianFabian
19.7k3674
$endgroup$
Hint:
Use $operatorname{Re} e^{ix} =cos x$. Then choose a contour that encircles the upper half-plane.
By using the residue theorem, we get the result
$$I= int_{-infty}^{infty} frac{cos x}{cosh^2 x},dx = operatorname{Re} sum_{k=0}^infty 2pi i,operatorname{Res}_{x=ipi (k+1/2)} left(frac{e^{ix}}{cosh^2 x}right).$$
We can calculate the residues as (the poles are of second order and $cosh^{-2}(x) approx -(x-x_0)^{-2}$ close to the pole at $x_0$.)
$$operatorname{Res}_{x=ipi (k+1/2)} left(frac{e^{ix}}{cosh^2 x}right) = - frac{d}{dx} e^{ix} Big|_{x=ipi (k+1/2)} = -i e^{-pi(k+1/2)},.$$
The integral then assumes the form
$$ I= 2pi sum_{k=0}^infty e^{-pi(k+1/2)} = frac{2pi e^{-pi/2} }{1-e^{-pi}} = frac{pi}{sinh(pi/2)},.$$
answered Jan 9 at 12:44
FabianFabian
19.7k3674
edited Jan 9 at 13:24
answered Jan 9 at 12:44
FabianFabian
19.7k3674
answered Jan 9 at 12:44
FabianFabian
19.7k3674
answered Jan 9 at 12:44
FabianFabian
19.7k3674
19.7k3674
$begingroup$
Aren't the poles of second order? The hyperbolic cosine function is squared.
$endgroup$
– Élio Pereira
Jan 9 at 13:15
1
$begingroup$
@ÉlioPereira: thanks, I will correct it.
$endgroup$
– Fabian
Jan 9 at 13:24
$begingroup$
Thank you very much for your help.
$endgroup$
– Élio Pereira
Jan 9 at 13:32
$begingroup$
Can you help me with just one more thing? How could I show that the integral on the arc is null?
$endgroup$
– Élio Pereira
Jan 9 at 16:32
1
$begingroup$
@ÉlioPereira: there are different ways to show that. The most straightforward way might be that $|e^{i z}/cosh(z)^2| = 2e^{-y}/(cos 2 y + cosh 2x) leq 2 e^{-y}/(cosh2 x -1)$ for $z=x+i y$. On the arc, we have that $yto+infty$ fast enough that the integral vanishes. To show this technically, you might have to distinguish points $z=r e^{itheta}$ with $|theta -pi/2| < epsilon$ from the rest.
$endgroup$
– Fabian
Jan 9 at 17:56
|
show 5 more comments
$begingroup$
Aren't the poles of second order? The hyperbolic cosine function is squared.
$endgroup$
– Élio Pereira
Jan 9 at 13:15
1
$begingroup$
@ÉlioPereira: thanks, I will correct it.
$endgroup$
– Fabian
Jan 9 at 13:24
$begingroup$
Thank you very much for your help.
$endgroup$
– Élio Pereira
Jan 9 at 13:32
$begingroup$
Can you help me with just one more thing? How could I show that the integral on the arc is null?
$endgroup$
– Élio Pereira
Jan 9 at 16:32
1
$begingroup$
@ÉlioPereira: there are different ways to show that. The most straightforward way might be that $|e^{i z}/cosh(z)^2| = 2e^{-y}/(cos 2 y + cosh 2x) leq 2 e^{-y}/(cosh2 x -1)$ for $z=x+i y$. On the arc, we have that $yto+infty$ fast enough that the integral vanishes. To show this technically, you might have to distinguish points $z=r e^{itheta}$ with $|theta -pi/2| < epsilon$ from the rest.
$endgroup$
– Fabian
Jan 9 at 17:56
$begingroup$
Aren't the poles of second order? The hyperbolic cosine function is squared.
$endgroup$
– Élio Pereira
Jan 9 at 13:15
$begingroup$
Aren't the poles of second order? The hyperbolic cosine function is squared.
$endgroup$
– Élio Pereira
Jan 9 at 13:15
1
1
$begingroup$
@ÉlioPereira: thanks, I will correct it.
$endgroup$
– Fabian
Jan 9 at 13:24
$begingroup$
@ÉlioPereira: thanks, I will correct it.
$endgroup$
– Fabian
Jan 9 at 13:24
$begingroup$
Thank you very much for your help.
$endgroup$
– Élio Pereira
Jan 9 at 13:32
$begingroup$
Thank you very much for your help.
$endgroup$
– Élio Pereira
Jan 9 at 13:32
$begingroup$
Can you help me with just one more thing? How could I show that the integral on the arc is null?
$endgroup$
– Élio Pereira
Jan 9 at 16:32
$begingroup$
Can you help me with just one more thing? How could I show that the integral on the arc is null?
$endgroup$
– Élio Pereira
Jan 9 at 16:32
1
1
$begingroup$
@ÉlioPereira: there are different ways to show that. The most straightforward way might be that $|e^{i z}/cosh(z)^2| = 2e^{-y}/(cos 2 y + cosh 2x) leq 2 e^{-y}/(cosh2 x -1)$ for $z=x+i y$. On the arc, we have that $yto+infty$ fast enough that the integral vanishes. To show this technically, you might have to distinguish points $z=r e^{itheta}$ with $|theta -pi/2| < epsilon$ from the rest.
$endgroup$
– Fabian
Jan 9 at 17:56
$begingroup$
@ÉlioPereira: there are different ways to show that. The most straightforward way might be that $|e^{i z}/cosh(z)^2| = 2e^{-y}/(cos 2 y + cosh 2x) leq 2 e^{-y}/(cosh2 x -1)$ for $z=x+i y$. On the arc, we have that $yto+infty$ fast enough that the integral vanishes. To show this technically, you might have to distinguish points $z=r e^{itheta}$ with $|theta -pi/2| < epsilon$ from the rest.
$endgroup$
– Fabian
Jan 9 at 17:56
|
show 5 more comments
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