# 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.

Please show full working.

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

• what substitution did you make? Aug 5, 2016 at 23:57
• You might receive a response sooner if you include your work so that others can provide suggestions.
– Em.
Aug 5, 2016 at 23:59
• Question updated Aug 6, 2016 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.) Aug 6, 2016 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. Aug 6, 2016 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? Aug 6, 2016 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? Aug 6, 2016 at 14:30