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A trigonometric log integral

Evaluate the integral

\displaystyle \mathcal{J}=\int_0^{\pi/2} \frac{\log (1 + \sin x)}{\sin x} \, {\rm d}x

Solution

We are invoking a differentiation under the integral sign but placing a parameter is rather tricky. Let us consider the function

\displaystyle f(a) = \int_{0}^{\pi/2} \frac{\log \left ( 1+ \cos a \sin x \right )}{\sin x} \, {\rm d}x

where a is imposed on the restriction 0\leq a \leq \pi. Applying the substitution u=\frac{\pi}{2} - x we have that

\displaystyle f(a) = \int_{0}^{\pi/2} \frac{\log \left ( 1+ \cos a \cos x \right )}{\cos x} \, {\rm d}x

Differentiating with respect to a we have that

    \begin{align*} f'(a) &= \frac{\mathrm{d} }{\mathrm{d} a}\int_{0}^{\pi/2} \frac{\log \left ( 1+ \cos a \cos x \right )}{\cos x} \, {\rm d}x \\ &= \int_{0}^{\pi/2} \frac{\partial }{\partial a} \frac{\log \left ( 1+ \cos a \cos x \right )}{\cos x} \, {\rm d}x\\ &=-\int_{0}^{\pi/2} \frac{\sin a}{1 + \cos a \cos x} \, {\rm d}x \\ &= \left [ -2 \arctan \left ( \arctan \frac{a}{2} \tan \frac{x}{2} \right ) \right ]_0^{\pi/2}\\ &= -a \end{align*}

We also note that f \left ( \frac{\pi}{2} \right ) = 0 . Thus, we conclude that

 \displaystyle f(a) = \frac{\pi^2}{8} - \frac{a^2}{2}

For a=0 we get that the initial integral is equal to

\displaystyle \int_0^{\pi/2} \frac{\log (1 + \sin x)}{\sin x} \, {\rm d}x = \frac{\pi^2}{8}

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