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Hi someone can help me with this integral, I can imagine that it is solvable with method of residues but in this case I do not know where to start. $$\frac{1}{\sqrt{2\pi}}\int_{-\infty }^{\infty }\frac{k}{2}\sqrt{\frac{\pi}{2}}e^{-2|k-\pi|}dk$$

Moreover without the absolute value of the following integrals do not converge $$\frac{1}{\sqrt{2\pi}}\int_{-\infty }^{\infty }\frac{k}{2}\sqrt{\frac{\pi}{2}}e^{-2(k-\pi)}dk$$ $$\frac{1}{\sqrt{2\pi}}\int_{-\infty }^{\infty }\frac{k}{2}\sqrt{\frac{\pi}{2}}e^{2(k-\pi)}dk$$ Thank you so much for your help

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    Split the integration region $(-\infty,\infty)$ into the $(-\infty,\pi]$ and $[\pi,\infty)$ for which $|k-\pi| = \pi - k$ and $k-\pi$ respectively. This gives in general $\int_{-\infty}^\infty g(k)f(|k-\pi|){\rm d}k = \int_{-\infty}^\pi g(k)f(\pi-k){\rm d}k + \int_{\pi}^\infty g(k)f(k-\pi){\rm d}k$ and both of these integrals exists in this case. – 2017-02-19
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    @Winther Hans, while the analysis in your comment is correct, that way forward is a bit inefficient. A simple substitution $k-\pi \to k$ and exploitation of symmetry renders evaluation trivial. ;-)) – 2017-02-19

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Let $I$ be given by the integral

$$I=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \frac k2 \sqrt{\frac{\pi}{2}}e^{-2|k-\pi|}\,dk\tag 1$$

Enforcing the substitution $k-\pi \to k$ in $(1)$ yields

$$I=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \frac {k+\pi}2 \sqrt{\frac{\pi}{2}}e^{-2|k|}\,dk \tag 2$$

Then, exploiting symmetry (i.e., $ke^{-2|k|}$ is odd and $e^{-2|k|}$ is even), we find that

$$I=\frac\pi2\int_0^\infty e^{-2k}\,dk=\frac{\pi}{4}$$

And we are done!