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I am very new to the Residuetheorem and now I am asked to calculate the following integral:

$$\int_0^\infty \frac{\cos(t)}{(1+t^2)^3}\text{d}t$$

I know it has poles of order $3$ at $x=\pm i$ and that I have to find a closed curve in order to calculate it.

But as I said, I am very new to this and (to be hounest) a little lost at the moment. Therefore any hint or help is very much appreciated!

José Carlos Santos
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TwoStones
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    Because the integrand is even, the integral is half of that evaluated along the entire real axis. Furthermore, we can say $\cos t = \mathrm{Re}[e^{it}]$ and apply Jordan's lemma. – Sohom Paul Jul 06 '19 at 16:37
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    Not residue based, but I posted a more generalised form (seeking Real Methods) yesterday and was fortunate to have a solution that was simple and elegant. https://math.stackexchange.com/questions/3284703/i-nt-a-int-0-infty-frac-cosxt-leftx2-a2-rightn-dx –  Jul 07 '19 at 08:14

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Expanding on what I put as my comment, the standard approach to this integral would be to realize that $$\int_{0}^\infty \frac{\cos t}{(1+t^2)^3}\, dt = \frac{1}{2}\textrm{Re}\left[\lim_{R\to\infty}\int_{\gamma(R)}\frac{e^{iz}}{(1+z^2)^3}\, dz\right]$$ where $\gamma(R)$ is the contour that goes from $-R$ to $R$ along the real axis and travels along a semicircle in the upper half-plane to complete the loop. Using Jordan's lemma, we can see that the upper semicircle's contribution to the integral vanishes in the limit. Our only residue in this contour is the one at $i$, which may be easily calculated to be $$\textrm{Res}\left[\frac{e^{iz}}{(1+z^2)^3}, i\right] = \frac{1}{2!}\lim_{z\to i}\frac{d^2}{dz^2}\frac{e^{iz}}{(z + i)^3} = \frac{1}{2}\lim_{z\to i}\left(\frac{-e^{iz}}{(z+i)^3} - \frac{6ie^{iz}}{(z+i)^4} + \frac{12e^{iz}}{(z+i)^5}\right) = \frac{7}{16ie}$$ Thus, our loop integral evaluates to $2\pi i\textrm{Res}(i) = \frac{7\pi}{8e}$, meaning our original integral is merely $\boxed{\frac{7\pi}{16e}}$

Sohom Paul
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Note that your integral is equal to$$\frac12\operatorname{Re}\left(\int_{-\infty}^\infty\frac{e^{it}}{(1+t^2)^3}\,\mathrm dt\right).$$And$$\int_{-\infty}^\infty\frac{e^{it}}{(1+t^2)^3}\,\mathrm dt=2\pi i\operatorname{res}_{z=i}\frac{e^{iz}}{(1+z^2)^3}.$$Finally, this last residue is equal to $-\frac{7i}{16e}$. Therefore, your integral is equal to $\frac{7\pi}{16e}$.

José Carlos Santos
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