8

Prove that $$\int_0^{\frac{\pi}{2}}\frac{\log\left(\sin x\right)}{\cos^2\left(x\right)+y^2\sin^2\left(x\right)}{\rm d}x=-\frac{\pi}{2}\frac{\log\left(1+y\right)}{y}$$ where $y\ge0$.

I came across this problem on a math forum, but after attempting to solve it, I didn't get any results. I attempted to solve the problem using the residue theorem but found it challenging to prove the expression. I believe the issue lies in how to handle $\log(\sin x)$, which might require some clever substitution. But I couldn't think of any.

I hope someone can provide the correct solution. Thanks!

Quanto
  • 120,125
displllau
  • 81
  • I know an integral $\int \log(1-2y \cos(x)+y^2) dx$ that uses a) differentiation under the integral acc. to $y,$ b) Weierstraß substitution (half tangent subst.), c) another substitution $\dfrac{1+y}{1-y}$ and finally d) $\int_0^{\pi/2}\log(\sin(x))dx=-(\pi/2)\log(2).$ It is rather tricky. It also has an easier solution if you use a series expansion. In my case, I found that expansion in Gradshteyn, Ryzhik 1.514, Another list where you could find ideas is https://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Bestimmte_Integrale (wrong language, but it's basically only formulas, so what) – Marius S.L. Apr 19 '24 at 13:31

2 Answers2

10

\begin{align} &\int_0^{\frac{\pi}{2}}\frac{\ln\left(\sin x\right)}{\cos^2x+y^2\sin^2x}{d}x\\ = & -\int_0^{\frac{\pi}{2}}\int_0^1 \frac{t\cot^2x}{(\cos^2x+y^2\sin^2x)(1+t^2\cot^2x)}dt\ dx\\ = & -\int_0^1\int_0^{\frac{\pi}{2}}\frac{t\sec^2x}{(1+y^2\tan^2x)(t^2+\tan^2x)}dx\ dt\\ =& -\frac\pi{2} \int_0^1 \frac1{ty+1}dt =-\frac{\pi}{2y}{\ln(1+y)} \end{align}

Quanto
  • 120,125
  • Would you mind to look at https://math.stackexchange.com/questions/4763428/is-it-possible-to-integrate-frac-tan-1tt2n-sqrtt2-1 which never received any answer ? Thanks in advance. I always admire your integrations. Cheers & thanks – Claude Leibovici Apr 19 '24 at 13:33
5

Use the following result Two integrals: $$\int_{0}^{\infty}\frac{\ln u}{1+y^2u^2}{\rm d}u=\frac{-\pi}{2y}\ln y,\quad \frac12\int_{0}^{\infty}\frac{\ln(1+u^2)}{1+y^2u^2}{\rm d}u =\frac{\pi}{2y}\ln\frac{y+1}{y},$$ Let $u=\tan x$, then $$\int_0^{\frac{\pi}{2}}\frac{\log\left(\sin x\right)}{\cos^2\left(x\right)+y^2\sin^2\left(x\right)}{\rm d}x =\frac12\int_{0}^{\infty}\frac{2\ln u-\ln(1+u^2)}{1+y^2u^2}{\rm d}u =-\frac{\pi}{2}\frac{\log\left(1+y\right)}{y}.$$

Riemann
  • 11,801