Sunday, December 5, 2021

Integral of the day XIII

Another fun integral

Integral involving $\zeta(2)$


Today we show the proof of this result posted by @integralsbot \[\int_{0}^{\pi} \frac{\ln\left(1+\frac{\cos x}{2}\right)}{\cos x} dx = \zeta(2)\]

Proof:

Consider the following integral: \[ \Phi(a) = \int_{0}^{\pi} \frac{\ln\left(1+a\frac{\cos x}{2}\right)}{\cos x} dx \quad 0\lt a\lt 2 \] Differentiating under the integral sing and using the fact that $\displaystyle \frac{1}{2\cos x + a }$ is an even function : \[ \Phi'(a) = \int_{0}^{\pi} \frac{1}{2\cos x + a } dx = \frac{1}{2}\int_{-\pi}^{\pi} \frac{1}{2\cos x + a } dx \] If we make the substitution: \[ \cos x = \frac{z+z^{-1}}{2} \] \[ dx = \frac{dz}{zi} \] we transform this integral in a contour integral round the unit complex circle \[ \Phi'(a) = \frac{1}{2}\int_{-\pi}^{\pi} \frac{1}{2\cos x + a } dx = -i\oint_{|z|=1} \frac{dz}{az^2+4z+a}\] The function \[ f(z) = \frac{1}{az^2+4z+a}\] has two poles at $\displaystyle z = \frac{\sqrt{4-a^2}-2}{a}$ and $\displaystyle z = \frac{-\sqrt{4-a^2}-2}{a} $

The condition that $a>0$ assures that the only pole inside the unit complex circle is $\displaystyle z = \frac{\sqrt{4-a^2}-2}{a}$

By the Cauchy integral formula:

\[ \Phi'(a) = -i\oint_{|z|=1} \frac{dz}{az^2+4z+a} = \frac{2\pi}{a\left(\frac{\sqrt{4-a^2}-2}{a} + \frac{\sqrt{4-a^2}+2}{a}\right)} = \frac{\pi}{\sqrt{4-a^2}}\] Integrating with respect to $a$: \begin{align*} \Phi(a) = \pi\int \frac{da}{\sqrt{4-a^2}} + C =& \frac{\pi}{2} \int \frac{da}{\sqrt{1-\frac{a^2}{4}}} + C\\ =& \pi \int \frac{dw}{\sqrt{1-w^2}} + C \quad \left(w^2 \mapsto \frac{a^2}{4}\right) \\ =& \pi\arcsin(w) + C\\ =& \pi\arcsin\left(\frac{a}{2}\right) + C \end{align*} Therefore \[ \Phi(a) = \int_{0}^{\pi} \frac{\ln\left(1+a\frac{\cos x}{2}\right)}{\cos x} dx = \pi\arcsin\left(\frac{a}{2}\right) + C\] If we let $a\to 0+$ then $\Phi(0)=0$, $\arcsin(0)=0$ and $C=0$

If we put $a=1$ \[ \Phi(1) = \int_{0}^{\pi} \frac{\ln\left(1+\frac{\cos x}{2}\right)}{\cos x} dx = \frac{\pi^2}{6} = \zeta(2)\] Hence, we can conclude \[\boxed{\int_{0}^{\pi} \frac{\ln\left(1+a\frac{\cos x}{2}\right)}{\cos x} dx = \pi\arcsin\left(\frac{a}{2}\right) \quad 0\lt a \lt 2} \] \[\boxed{\int_{0}^{\pi} \frac{\ln\left(1+\frac{\cos x}{2}\right)}{\cos x} dx = \zeta(2)}\]

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