Evaluate $\int_0^{\pi/4} \ln\left(\ln\left(\cot\left(x\right)\right)\right) {d}x$

Asked Dec 12 '20 at 14:44

Active Dec 12 '20 at 15:04

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I've seen this integral on a math page : $$\int_0^{\pi/4} \ln\left(\ln\left(\cot\left(x\right)\right)\right) {d}x.$$ And I tried to solve it, like Vardi's integral, by substituting $u=\ln\left(\cot(x)\right)$

The bounds are : $u\left(\frac{\pi}{4}\right)=\ln\left(\cot\left( \frac{\pi}{4}\right)\right)=\ln(1)=0$ and $u(0)=\ln\left(\cot\left( 0\right)\right)=+\infty$ and this explains the minus sign we'll have it when we'll substitute $$u=\ln\left(\cot(x)\right) \Longleftrightarrow du=-\frac{dx}{\cos x\sin x}.$$

The problem now is just to simplify this expression I did a lot of attempts but I didn't get something ! I think that the substitution is false, any tips or thoughts ?

edited Dec 12 '20 at 15:04

GNUSupporter 8964民主女神地下教會

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asked Dec 12 '20 at 14:44

euler_med

1,044

1

Please don't use \frac in exponents or limits of integrals. It looks bad and confusing, and it rarely appears in professional mathematics typesetting. – GNUSupporter 8964民主女神地下教會 Dec 12 '20 at 14:56
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@GNUSupporter8964民主女神地下教會 Sure, I'll edit it right now ! Thanks for the advice tho :D – euler_med Dec 12 '20 at 14:57
@Angelo Your edit collided with OP's one. – GNUSupporter 8964民主女神地下教會 Dec 12 '20 at 15:04
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You substitution leads to solve another very difficult integral $$\int_0^{\infty} \frac{1}{2} \log (u) \text{sech}(u) , du=\frac{1}{4} \pi \log \left(\frac{4 \pi ^3}{\Gamma \left(\frac{1}{4}\right)^4}\right)$$ – Raffaele Dec 12 '20 at 15:05
@GNUSupporter8964民主女神地下教會 It doesn't :'( I tried the same steps with more intuition and I didn't get a result !! – euler_med Dec 12 '20 at 15:07
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Isn't this integral same as the Vardi's integral after substituting $y = \pi/2 -x $? – sudeep5221 Dec 12 '20 at 15:10
This seems useful so I'll link it here. (https://math.stackexchange.com/questions/1585960/improper-integral-of-log-x-operatornamesech-x) – Khallil Dec 12 '20 at 15:16

Evaluate $\int_0^{\pi/4} \ln\left(\ln\left(\cot\left(x\right)\right)\right) {d}x$

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