1. ## contour integration

My professor has asked us to show that, for $m>0$,

$
\int_{0}^{\infty}dx\frac{\cos[mx]}{\left(x^{2}+1\right)^{2}}=\frac{\pi}{4}\left(1+m \right)\exp[-m]
$

using the Residue theorem (in this case:

$
\oint_Cdz\frac{\cos[mz]}{(z^2+1)^2}=\pi i\mathrm{Res}\left[\frac{\cos[mz]}{(z^2+1)^2},\,i\right]
$

My solution is

$
\mathrm{Res}\left[\frac{\cos[mz]}{\left(z+i\right)^{2}\left(z-i\right)^{2}},\, i\right]=\frac{d}{dz}\left((z-i)^{2}\frac{\exp[imz]}{(z+i)^{2}(z-i)^{2}}\right)_{z=i}
$

$
=\frac{d}{dz}\left(\frac{\exp[imz]}{(z+i)^{2}}\right)_{z=i}
$

$
=\left.\frac{im\exp[imz]}{(z+i)^{2}}-\frac{2\exp[imz]}{(z+i)^{3}}\right|_{z=i}
$

$
=\frac{im\exp[-m]}{4i^{2}}-\frac{2\exp[-m]}{8i^{3}}
$

$
=\frac{m\exp[-m]}{4i}+\frac{\exp[-m]}{4i}=\frac{1}{4i}(m+1)\exp[-m]
$

$
\therefore\int_0^\infty dx\frac{\cos[mx]}{(x^2+1)^2}=\frac{\pi}{4}(m+1)\exp[-m]
$

I get the answer no problem, but I'm not sure about is why this is for $m>0$ only. If I let $m<0$, I get (by the same/similar arguments and mathematics as above)

$
\int_0^\infty dx\frac{\cos[mx]}{(x^2+1)^2}=\frac{\pi}{4}(1-m)\exp[m]
$

And I'm not sure why this wouldn't be allowed? Any suggestions of where to look?

2. Originally Posted by jdwood983
My professor has asked us to show that, for $m>0$,

$
\int_{0}^{\infty}dx\frac{\cos[mx]}{\left(x^{2}+1\right)^{2}}=\frac{\pi}{4}\left(1+m \right)\exp[-m]
$

using the Residue theorem (in this case:

$
\oint_Cdz\frac{\cos[mz]}{(z^2+1)^2}=\pi i\mathrm{Res}\left[\frac{\cos[mz]}{(z^2+1)^2},\,i\right]
$

My solution is

$
\mathrm{Res}\left[\frac{\cos[mz]}{\left(z+i\right)^{2}\left(z-i\right)^{2}},\, i\right]=\frac{d}{dz}\left((z-i)^{2}\frac{\exp[imz]}{(z+i)^{2}(z-i)^{2}}\right)_{z=i}
$

$
=\frac{d}{dz}\left(\frac{\exp[imz]}{(z+i)^{2}}\right)_{z=i}
$

$
=\left.\frac{im\exp[imz]}{(z+i)^{2}}-\frac{2\exp[imz]}{(z+i)^{3}}\right|_{z=i}
$

$
=\frac{im\exp[-m]}{4i^{2}}-\frac{2\exp[-m]}{8i^{3}}
$

$
=\frac{m\exp[-m]}{4i}+\frac{\exp[-m]}{4i}=\frac{1}{4i}(m+1)\exp[-m]
$

$
\therefore\int_0^\infty dx\frac{\cos[mx]}{(x^2+1)^2}=\frac{\pi}{4}(m+1)\exp[-m]
$

I get the answer no problem, but I'm not sure about is why this is for $m>0$ only. If I let $m<0$, I get (by the same/similar arguments and mathematics as above)

$
\int_0^\infty dx\frac{\cos[mx]}{(x^2+1)^2}=\frac{\pi}{4}(1-m)\exp[m]
$

And I'm not sure why this wouldn't be allowed? Any suggestions of where to look?

It's exactly the same, of course, as $\cos x = \cos (-x)\,,\,\,\forall\,x\in \mathbb{R}$.

Tonio

Tonio

3. So it's not that it's not allowed, it's that it's the same regardless. Seems strange then that the professor would call attention to a seemingly moot point.

4. I get (for $m\in\mathbb{R}$):


\begin{aligned}
\int_0^{\infty}\frac{\cos(mx)}{(x^2+1)^2}dx&=\frac {1}{2}\mathop\oint\limits_{D_u} \frac{\cos(mz)}{(z^2+1)^2}dz \\
&=\frac{1}{2}\left(2\pi i\mathop\text{Res}\limits_{z=i} \frac{\cos(mz)}{(z^2+1)^2}\right)\\
&=\frac{1}{2}2\pi i\big(-\frac{i}{4}(\cosh(m)-m\sinh(m))\big)\\
&=\frac{\pi}{4}(\cosh(m)-m\sinh(m))
\end{aligned}

5. Originally Posted by shawsend
I get (for $m\in\mathbb{R}$):


\begin{aligned}
\int_0^{\infty}\frac{\cos(mx)}{(x^2+1)^2}dx&=\frac {1}{2}\mathop\oint\limits_{D_u} \frac{\cos(mz)}{(z^2+1)^2}dz \\
&=\frac{1}{2}\left(2\pi i\mathop\text{Res}\limits_{z=i} \frac{\cos(mz)}{(z^2+1)^2}\right)\\
&=\frac{1}{2}2\pi i\big(-\frac{i}{4}(\cosh(m)-m\sinh(m))\big)\\
&=\frac{\pi}{4}(\cosh(m)-m\sinh(m))
\end{aligned}
I made an assumption that I didn't write down; so here is the corrected first line:

$

\mathrm{Res}\left[\frac{\cos[mz]}{\left(z+i\right)^{2}\left(z-i\right)^{2}},\, i\right]=\frac{d}{dz}\left((z-i)^{2}\frac{\Re[\exp[imz]]}{(z+i)^{2}(z-i)^{2}}\right)_{z=i}
$

Doing this gets you the right answer