I stumbled upon this problem which I cant seem to solve :

Prove that :

$$\int_0^\frac{\pi}{2}\frac{\log(1+\cos\alpha \cos x)}{\cos x}dx=\frac{1}{2}\left[\frac{\pi^2}{4}-\alpha^2\right]$$

I've tried all that i could.
Any solution for this?
If not a solution, maybe a hint or a direction in which i could think?

  • 1
    $\begingroup$ Hint: Can you differentiate on $\alpha$ under the integration sign? Can you justify it? $\endgroup$ – user491874 Dec 31 '17 at 22:06

Let $$I (\alpha) = \int_0^{\pi/2} \frac{\ln (1 + \cos \alpha \cos x)}{\cos x} \, dx, \quad \alpha \in \mathbb{R}.$$ Note that $I(\frac{\pi}{2}) = 0$. Now differentiating with respect to the parameter $\alpha$ we have $$I'(\alpha) = - \sin \alpha \int_0^{\pi/2} \frac{dx}{1 + \cos \alpha \cos x}.$$ The resulting integral can be evaluated using a Weierstrass substitution of $t = \tan (x/2)$. Doing so yields $$I'(\alpha) = - 2\sin \alpha \int_0^1 \frac{dt}{1 + \cos \alpha + (1 - \cos \alpha) t^2}.$$

Now as the term $1 - \cos \alpha \geqslant 0$ for all $\alpha \in \mathbb{R}$ we can write \begin{align*} I'(\alpha) &= -\frac{2 \sin \alpha}{1 - \cos \alpha} \int_0^1 \frac{dt}{ \left (\sqrt{\frac{1 + \cos \alpha}{1 - \cos \alpha}} \right )^2 + t^2}\\ &= -\frac{2 \sin \alpha}{1 - \cos \alpha} \sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} \tan^{-1} \left (\sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} \right )\\ &= -\alpha, \end{align*} where in the last line the following well-known result for the inverse tangent function has been used $$\cos^{-1} (x) = 2 \tan^{-1} \left (\sqrt{\frac{1 - x}{1 + x}} \right ), \quad -1 < x \leqslant 1.$$

So on integrating up with respect to $\alpha$ we have $$I(\alpha) = -\frac{\alpha^2}{2} + C,$$ where $C$ is an arbitrary constant to be determined. Since $I(\frac{\pi}{2}) = 0$, one finds $C = \pi^2/8$. Thus $$I(\alpha) = \frac{1}{2} \left (\frac{\pi^2}{4} - \alpha^2 \right ),$$ as required.


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