# Help with second integral in a Cauchy's integral formula problem.

I have been trying to do this problem for a while:

Use Cauchy's integral formula to evaluate $$\int_{-\infty}^\infty \frac{t\operatorname{sin}(\pi t)}{t^2+4}dt.$$

I have factored it into $$\int_{-\infty}^\infty \frac{t\operatorname{sin}(\pi t)}{t^2+4}dt=\frac{1}{2i}\left(\int_{-\infty}^\infty \frac{te^{i\pi t}}{t^2+4}dt-\int_{-\infty}^\infty \frac{te^{-i\pi t}}{t^2+4}dt\right).$$

So for first integral I am supposed to split it up into $\oint f dz - \int_{\gamma}f dz$ where $f$ is the integrand above and $\Gamma$ is a circle of radius $R$ in the upper half plane (ie $\gamma(t)=Re^{i\theta}:0\leq\theta<\pi$). But I can't seem to evaluate the second integral in this formula - the $\int_{\gamma}f dz$.

I'm sure this is obvious but I could use some help.

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$\displaystyle \int_{-\infty}^{\infty} \frac{t \sin \pi t}{t^{2}+4} \ dt = \text{Im} \int_{-\infty}^{\infty} \frac{t e^{i \pi t}}{t^{2}+4} \ dt$ –  Random Variable Feb 20 '13 at 16:01
Follow Random Variable's hint. Then you only need to do one integral. If you absolutely want to do both, you need to use a semicircle in the lower half plane for the other one. (Remember the orientation!) –  mrf Feb 20 '13 at 16:15
First, consider the complex integral $$\int_{C} \frac{z\operatorname{e}^{i\pi z}}{z^2+4}dz,$$ then use Random Variable hint. –  Mhenni Benghorbal Feb 20 '13 at 16:26
By the way, $z=x+iy$ and for the contour $C$, see here. –  Mhenni Benghorbal Feb 20 '13 at 16:48

\begin{align} \int_{-\infty}^\infty\frac{t\sin(\pi t)}{t^2+4}dt &=\frac1{2i}\int_{-\infty}^\infty\frac{te^{i\pi t}}{t^2+4}\mathrm{d}t -\frac1{2i}\int_{-\infty}^\infty\frac{te^{-i\pi t}}{t^2+4}\mathrm{d}t\tag{1}\\ &=\frac1{2i}\int_{\gamma^+}\frac{te^{i\pi t}}{t^2+4}\mathrm{d}t -\frac1{2i}\int_{\gamma^-}\frac{te^{-i\pi t}}{t^2+4}\mathrm{d}t\tag{2}\\ &=\frac{2\pi i}{2i}\left(\frac{2ie^{-2\pi}}{4i}\right) +\frac{2\pi i}{2i}\left(\frac{-2ie^{-2\pi}}{-4i}\right)\tag{3}\\ &=\pi e^{-2\pi}\tag{4} \end{align} $(1)\quad\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$
$(2)\quad\gamma^+$ follows the real axis and circles counterclockwise back through the upper half plane.
$\hphantom{(2)}\quad\gamma^-$ follows the real axis and circles clockwise back through the lower half plane.
$(3)\quad$evaluate the residues at $+2i$ and $-2i$ and note that $\gamma^-$ is clockwise.
We choose $\gamma^+$ for $e^{i\pi t}$ since $e^{i\pi t}$ decays in the upper half plane. Similarly, $e^{-i\pi t}$ decays in the lower half plane.