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## Homework Statement

Show that, for [itex]m>0[/itex],

[tex]

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

[/tex]

## Homework Equations

Residue theorem:

[tex]

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

[/tex]

## The Attempt at a Solution

[tex]

\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}

[/tex]

[tex]

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

[/tex]

[tex]

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

[/tex]

[tex]

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

[/tex]

[tex]

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

[/tex]

[tex]

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

[/tex]

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

[tex]

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

[/tex]

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

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