# How to Integrate $\int_0^\pi \ln(1+\alpha \cos(x)) \,\mathrm{d}x$

I've been trying to learn how to integrate by differentiation under the integral. I've made good progress on some problems, but I seem to not be able to get an answer for $$f(\alpha)=\int_0^\pi \ln(1+\alpha \cos(x)) \,\mathrm{d}x$$

I've managed to get as far as $$f'(\alpha)=\int_0^\pi \frac{\cos(x)}{1+ \alpha \cos(x)} \,\mathrm{d}x$$

But this seems like a ridiculous integral to try and integrate by elementary methods, indeed an integral calculator returns $$\dfrac{x}{a}+\dfrac{\ln\left(\left|\left(a-1\right)\tan\left(\frac{x}{2}\right)-\sqrt{a^2-1}\right|\right)-\ln\left(\left|\left(a-1\right)\tan\left(\frac{x}{2}\right)+\sqrt{a^2-1}\right|\right)}{a\sqrt{a^2-1}}$$

Hopefully someone can advise on whether I've already made a mistake in my working, or whether I've just completely misunderstood the method.

• Now plug in the limits of integration. – Rene Schipperus Oct 6 '16 at 18:20
• One way is to convert the integral into a contour integral. This gives $$f'(\alpha) = \frac{\pi}{\alpha}\left( 1 - \frac{1}{\sqrt{1-\alpha^2}} \right)$$ when $|\alpha| < 1$. Another way is to expand the logarithm using the Taylor series, and when $|\alpha| < 1$ this leads to $$f(\alpha) = -\pi \sum_{n=1}^{\infty} \binom{-1/2}{n} \frac{\alpha^{2n}}{n} = \int_{0}^{\alpha} \frac{\pi}{t} \left( 1 - \frac{1}{\sqrt{1-t^2}} \right) \, dt.$$ Both way gives $f(\alpha ) = \pi \log \left( \frac{ 1 + \sqrt{1-\alpha^2}}{2} \right)$. – Sangchul Lee Oct 6 '16 at 18:55
• Possible duplicate of Verification this integral identity – Nosrati Aug 27 '18 at 5:53

Before anything else, note that $$f(0)=0$$, so our constant of integration will be $$0$$.
As you found, $$f'(a)=\int_0^\pi\frac{\cos x}{1+a\cos x}dx$$ $$f'(a)=\frac1a\int_0^\pi\frac{1-1+a\cos x}{1+a\cos x}dx$$ $$f'(a)=\frac1a\int_0^\pi\frac{1+a\cos x}{1+a\cos x}dx-\frac1a\int_0^\pi\frac{dx}{1+a\cos x}$$ $$f'(a)=\frac\pi a-\frac1aj(a)$$ For $$j$$, we preform $$t=\tan\frac{x}2$$ which gives $$j(a)=2\int_0^\infty \frac{1}{1+a\frac{1-t^2}{1+t^2}}\frac{dt}{1+t^2}$$ $$j(a)=\frac2{1-a}\int_0^\infty \frac{dt}{t^2+\frac{1+a}{1-a}}$$ It is easily shown that $$\int_0^\infty \frac{dx}{x^2+c}=\frac\pi{2\sqrt c}$$ So we have that $$j(a)=\frac\pi{(1-a)\sqrt{\frac{1+a}{1-a}}}=\frac\pi{\sqrt{1-a^2}}$$ Hence $$f'(a)=\frac\pi a-\frac\pi{a\sqrt{1-a^2}}$$ So $$f(a)=\pi\log|a|-\pi\int\frac{da}{a\sqrt{1-a^2}}$$ For the remaining integral, let $$a=\sin x$$ to get $$\int\frac{da}{a\sqrt{1-a^2}}=\int\frac{dx}{\sin x}=-\log\left(\cot x+\csc x\right)=-\log\frac{1+\sqrt{1-a^2}}{a}$$ So $$f(a)=\pi\log a+\pi\log\frac{1+\sqrt{1-a^2}}{a}$$ Which is just $$f(a)=\pi\log\left(1+\sqrt{1-a^2}\right)$$