12

Evaluate

$$\displaystyle \int^{\frac{\pi}{2}}_{0} x \ln(\cos x) \sqrt{\tan x}dx$$

Unfortunately, I have no idea on how to integrate this and thus cannot provide any inputs on my own. The only thing I noticed was that suppose we just had to find $$\int_0^{\pi/2} \ln(\sin(x))\ln(\cos(x)) dx$$ we could have written $$I(a,b)=\int_0^{\pi/2}\sin^a(x)cos^b(x)dx$$ converted this into the Beta and then the Gamma Function, taken its partial derivatives with respect to $a$ and $b$, and finally plugged in $a=b=0$ to get our answer. $$$$ I would be truly grateful if somebody could please help me solve this Integral. Many thanks!

epimorphic
  • 3,219
User1234
  • 3,958
  • Well, the fact that neither $\log{(\cos x)}$ or $\tan x$ is defined for $x=\dfrac{\pi}{2}$ is a problem, isn't it? – Martigan Jun 10 '15 at 12:14
  • @Martigan: no, it isn't. – Jack D'Aurizio Jun 10 '15 at 12:18
  • Sir, but since they are point discontinuities (I'm not sure about this), do they matter? It's just that @RolfHoyer Sir had told me that point discontinuities do not matter since the contribution of a point to an integral is 0. – User1234 Jun 10 '15 at 12:22
  • 1
    Few ideas: 1) complex contour integration maybe; 2) $\ln(\cos x)$ is an integral of a tangent, so integration by parts could actually help. Or a $\tan$-related substitution at least. – orion Jun 10 '15 at 12:25
  • you may find some ideas here: http://math.stackexchange.com/questions/1164183/how-to-evaluate-i-int-0-pi-2-fracx-log-sinx-sinx-dx/1172031#1172031 – tired Jun 10 '15 at 12:34
  • @orion Sir, I haven't learnt Contour Integrals. Sir, I cannot really understand how to use your other ideas. Please could you give a hint, Sir? – User1234 Jun 10 '15 at 12:36
  • By the way, Wolfram Alpha does not provide any closed form. – Crostul Jun 10 '15 at 12:38
  • is there any reason to believe that there is a closed form solution? – tired Jun 10 '15 at 12:41
  • See the close form https://math.stackexchange.com/a/4592534/686284 – Quanto Dec 06 '22 at 20:22

2 Answers2

5

We may compute the Fourier series of $\sqrt{\left|\tan\frac{x}{2}\right|}$ and $ \log\cos\frac{x}{2}$ over $I=(-\pi,\pi)$, for first.

We have: $$ \frac{1}{\pi}\int_{0}^{\pi}\sqrt{\left|\tan\frac{x}{2}\right|}\,dx = \sqrt{2},$$ $$ \frac{1}{\pi}\int_{-\pi}^{\pi}\cos(2mx)\sqrt{\left|\tan\frac{x}{2}\right|}\,dx=-\frac{1}{\pi}\int_{-\pi}^{\pi}\cos((2m+1)x)\sqrt{\left|\tan\frac{x}{2}\right|}\,dx=\frac{\sqrt{2}}{4^n}\binom{2m}{m}$$ from which:

$$\sqrt{\left|\tan\frac{x}{2}\right|} = \sum_{m\geq 0}\frac{\sqrt{2}}{4^m}\binom{2m}{m}\left[\cos(2mx)-\cos((2m+1)x)\right]\tag{1}$$

follows. On the other hand: $$ \frac{1}{2\pi}\int_{-\pi}^{\pi}\log\cos\frac{x}{2}\,dx = -\log 2, $$ $$ \frac{1}{\pi}\int_{-\pi}^{\pi}\log\left(\cos\frac{x}{2}\right)\cos(mx)\,dx = \frac{(-1)^{m+1}}{m},$$ hence:

$$ \log\cos\frac{x}{2}=-\log 2-\sum_{m=1}^{+\infty}\frac{(-1)^{m}}{m}\cos(mx)\tag{2} $$

so by using

$$ \int_{0}^{\pi} x\cos(nx)\cos(mx)\,dx = \frac{(-1)^{m+n}-1}{2}\left(\frac{1}{(m-n)^2}+\frac{1}{(m+n)^2}\right),\tag{3}$$ $$ \int_{0}^{\pi} x\cos^2(nx)\,dx = \frac{\pi^2}{4}\tag{4}$$

we get that our integral equals a rather complicated series.

Update: I managed to prove, through differentiation under the integral sign and the residue theorem, that our integral depends only on the values of $\psi(z)$ and $\psi'(z)$ at $z=\frac{1}{4}$ (see this related question). Both values are not too difficult to compute, and $\psi'\left(\frac{1}{4}\right)=\pi^2+8K$ where $K$ is the Catalan constant.

Community
  • 1
Jack D'Aurizio
  • 353,855
1

$$ I=-\frac{\pi}{8}\sqrt{2}({\pi}\ln2+4G+\frac{\pi^2}{3}+3\ln^22)$$ Where G is Catalan's constant.

wythagoras
  • 25,026
user178256
  • 5,467