# Prove via differentiation that integral $\int_0^\pi \frac{\log(1+\cos\alpha\cos\theta)}{\cos\theta}\,d\theta = \pi(\frac{\pi}{2}-\alpha)$

Prove via differentiation that integral $$\int_0^\pi \frac{\log(1+\cos\alpha\cos\theta)}{\cos\theta}\,d\theta$$ is equal to $$\pi\left(\frac{\pi}{2} - \alpha\right)$$ where $0\leq \alpha\leq \frac{\pi}{2}.$

After partially differentiating wrt $\alpha$ I have tried via substitution;

$$u=\log(1+\cos\alpha\cos\theta)$$

Then I get;

$$\frac{dI(\alpha)}{d\alpha}=-\sin\alpha\int_0^\pi \frac{1}{1+\cos\alpha\cos\theta}\,d\theta$$

I can not currently see how to progress this to stated answer.

The question is from G Stephenson's 'Mathematical Methods for Science students' - page 172.

• what substitution did you make? – астон вілла олоф мэллбэрг Aug 5 '16 at 23:57
• You might receive a response sooner if you include your work so that others can provide suggestions. – Em. Aug 5 '16 at 23:59
• Question updated – unseen_rider Aug 6 '16 at 0:16
• In principle, you can differentiate with respect to $\alpha$ once more. In that case the challenge is to show that the resulting definite integral vanishes identically. (I don't know if this is easier than the stated question.) – Semiclassical Aug 6 '16 at 0:37
• Also, what methods are you interested in? A proof via an appropriate complex contour integral is straightforward, for instance, but is not a real-analysis approach. – Semiclassical Aug 6 '16 at 0:41

Let $I(\alpha)$ be the integral of interest given by

$$I(\alpha)=\int_0^\pi \frac{\log(1+\cos(\alpha)\cos(\theta))}{\cos(\theta)}\,d\theta$$

Differentiating $I(\alpha)$, we obtain

$$\frac{dI(\alpha)}{d\alpha}=-\sin(\alpha)\int_0^\pi\frac{1}{1+\cos(\alpha)\cos(\theta)}\,d\theta$$

Note that we can use the classical Tangent Half-Angle Substitution to evaluate the integral

$$f(\alpha)=\int_0^\pi \frac{1}{1+\cos(\alpha)\cos(\theta)}\,d\theta=\frac{\pi}{\sin(\alpha)}$$

Therefore, we have $\frac{dI(\alpha)}{d\alpha}=-\pi$ whereupon integrating yields

$$I(\alpha)=-\pi \alpha +C$$

Noting that $I(\pi/2)=0$ reveals

$$I(\alpha)=\pi(\pi/2-\alpha)$$

as was to be shown!

NOTE:

To arrive at the closed-form expression for $f(\alpha)$ we enforce the substitution $t=\tan(x/2)$. Then, $\cos(x)=\frac{1-t^2}{1+t^2}$, $dx=\frac{2}{1+t^2}\,dt$, and the limits extend from $t=0$ to $t=\infty$ to reveal

\begin{align} f(\alpha)&=\frac{1}{\sin^2(\alpha/2)}\int_0^\infty \frac{1}{\cot^2(\alpha/2)+t^2}\,dt\\\\ &=\frac{1}{\sin^2(\alpha/2)}\frac{1}{\cot(\alpha/2)}\left.\left(\arctan\left(\frac{t}{\cot(\alpha/2)}\right)\right)\right|_{0}^{\infty}\\\\ &=\frac{1}{\frac12\sin(\alpha)} \,\frac{\pi}{2}\\\\ &=\frac{\pi}{\sin(\alpha)} \end{align}

• Please show further working on how you got $\cot^2(\frac{\alpha}{2}) + t^2$ in the denominator of the integral? – unseen_rider Aug 6 '16 at 12:27
• @unseen_rider Use the identities $1+\cos(x)=2\cos^2(x/2)$ and $1-\cos(x)=2\sin^2(x/2)$. Do you see now? – Mark Viola Aug 6 '16 at 14:30