# How to prove this sum of integrals

How to prove this ? $$\displaystyle \sum\limits_{k=0}^{\infty }{\int_{2k\pi }^{\left( 2k+1 \right)\pi }{{{\text{e}}^{-\frac{x}{2}}}\frac{\left| \sin x-\cos x \right|}{\sqrt{\sin x}}}}\text{d}x=\frac{2\cdot \sqrt[4]{8}{{\text{e}}^{-\frac{\pi }{8}}}}{1-{{\text{e}}^{-\pi }}}$$

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Any attempts from your side? – Eckhard Dec 29 '12 at 15:42

In the summand integral, perform a change of variables, $x= 2 \pi k + t$: $$\int_{2 \pi k}^{2 \pi k + \pi} \exp\left(-\frac{x}{2}\right) \frac{| \sin x - \cos x|}{\sqrt{ \sin x}} \mathrm{d}x = \exp(-\pi k) \int_0^\pi \exp(-t/2) \frac{| \sin t - \cos t|}{\sqrt{ \sin t}} \mathrm{d}t$$ Therefore: $$\sum_{k=1}^\infty \int_{2 \pi k}^{2 \pi k + \pi} \exp\left(-\frac{x}{2}\right) \frac{| \sin x - \cos x|}{\sqrt{ \sin x}} \mathrm{d}x = \sum_{k=0}^\infty \exp(-\pi k) \cdot \int_0^\pi \exp(-t/2) \frac{| \sin t - \cos t|}{\sqrt{ \sin t}} \mathrm{d}t$$ The problem is now split into two simpler problems: a trivial sum, and less trivial integral. The integral admits an anti-derivative: $$\frac{\mathrm{d}}{\mathrm{d}t} \left( 2 \sqrt{\sin t} \exp(-t/2) \right) = \exp(-t/2) \frac{\cos t - \sin t}{\sqrt{\sin t}}$$ You should be able to finish it off now.