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Math Help - contour integration

  1. #1
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    contour integration

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

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

    using the Residue theorem (in this case:

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

    My solution is

    <br />
\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}<br />

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

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

    <br />
=\frac{im\exp[-m]}{4i^{2}}-\frac{2\exp[-m]}{8i^{3}}<br />

    <br />
=\frac{m\exp[-m]}{4i}+\frac{\exp[-m]}{4i}=\frac{1}{4i}(m+1)\exp[-m]<br />

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

    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)

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

    And I'm not sure why this wouldn't be allowed? Any suggestions of where to look?
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  2. #2
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    Quote Originally Posted by jdwood983 View Post
    My professor has asked us to show that, for m>0,

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

    using the Residue theorem (in this case:

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

    My solution is

    <br />
\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}<br />

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

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

    <br />
=\frac{im\exp[-m]}{4i^{2}}-\frac{2\exp[-m]}{8i^{3}}<br />

    <br />
=\frac{m\exp[-m]}{4i}+\frac{\exp[-m]}{4i}=\frac{1}{4i}(m+1)\exp[-m]<br />

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

    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)

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

    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
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  3. #3
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    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.

    Thanks for your help!
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  4. #4
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    I get (for m\in\mathbb{R}):

    <br />
\begin{aligned}<br />
\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 \\<br />
&=\frac{1}{2}\left(2\pi i\mathop\text{Res}\limits_{z=i} \frac{\cos(mz)}{(z^2+1)^2}\right)\\<br />
&=\frac{1}{2}2\pi i\big(-\frac{i}{4}(\cosh(m)-m\sinh(m))\big)\\<br />
&=\frac{\pi}{4}(\cosh(m)-m\sinh(m))<br />
\end{aligned}<br />
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  5. #5
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    Quote Originally Posted by shawsend View Post
    I get (for m\in\mathbb{R}):

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

    <br /> <br />
\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}<br />

    Doing this gets you the right answer
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