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#### Pranav

##### Well-known member

- Nov 4, 2013

- 428

Compute:

$$\int_0^{\pi/2} \tan(x)\ln(\sin(x))\,dx$$

$$\int_0^{\pi/2} \tan(x)\ln(\sin(x))\,dx$$

- Thread starter Pranav
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- Thread starter
- #1

- Nov 4, 2013

- 428

Compute:

$$\int_0^{\pi/2} \tan(x)\ln(\sin(x))\,dx$$

$$\int_0^{\pi/2} \tan(x)\ln(\sin(x))\,dx$$

- Jan 17, 2013

- 1,667

$$\int^{\frac{\pi}{2}}_0 \sin^{2x-1}(\phi)\cos^{2y-1}(\phi)\, d\phi = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$

Hence we have by differentiation w.r.t to $x$

$$\int^{\frac{\pi}{2}}_0 \sin^{2x-1}(\phi)\cos^{2y-1}(\phi)\, \log(\sin(\phi)) d\phi = \frac{\Gamma(x)\Gamma(y)(\psi_0(x)-\psi_0(x+y))}{\Gamma(x+y)}$$

Putting \(\displaystyle x=1\) we have

$$\int^{\frac{\pi}{2}}_0 \sin(\phi)\cos^{2y-1}(\phi)\, \log(\sin(\phi)) d\phi = \frac{\Gamma(y)(\psi_0(1)-\psi_0(1+y))}{\Gamma(1+y)} = \frac{\psi_0(1)-\psi_0(1+y)}{y}$$

where I used that \(\displaystyle \Gamma (1+z)= z\Gamma(z)\)

Taking the limit as $y \to 0 $ we have

$$\int^{\frac{\pi}{2}}_0 \tan(\phi) \log(\sin(\phi)) d\phi= -\lim_{y \to 0}\frac{\psi_0(1+y)-\psi_0(1)}{y} = - \psi_1(1)=\frac{-\pi^2}{24}$$

Since \(\displaystyle \frac{d}{dz}\psi_0(z)=\psi_1(z)=\sum_{n\geq 1}\frac{1}{(n+z)^2}\)

- Thread starter
- #3

- Nov 4, 2013

- 428

Thanks ZaidAlyafey for your participation, your answer is correct.

$$\int^{\frac{\pi}{2}}_0 \sin^{2x-1}(\phi)\cos^{2y-1}(\phi)\, d\phi = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$

Hence we have by differentiation w.r.t to $x$

$$\int^{\frac{\pi}{2}}_0 \sin^{2x-1}(\phi)\cos^{2y-1}(\phi)\, \log(\sin(\phi)) d\phi = \frac{\Gamma(x)\Gamma(y)(\psi_0(x)-\psi_0(x+y))}{\Gamma(x+y)}$$

Putting \(\displaystyle x=1\) we have

$$\int^{\frac{\pi}{2}}_0 \sin(\phi)\cos^{2y-1}(\phi)\, \log(\sin(\phi)) d\phi = \frac{\Gamma(y)(\psi_0(1)-\psi_0(1+y))}{\Gamma(1+y)} = \frac{\psi_0(1)-\psi_0(1+y)}{y}$$

where I used that \(\displaystyle \Gamma (1+z)= z\Gamma(z)\)

Taking the limit as $y \to 0 $ we have

$$\int^{\frac{\pi}{2}}_0 \tan(\phi) \log(\sin(\phi)) d\phi= -\lim_{y \to 0}\frac{\psi_0(1+y)-\psi_0(1)}{y} = - \psi_1(1)=\frac{-\pi^2}{24}$$

Since \(\displaystyle \frac{d}{dz}\psi_0(z)=\psi_1(z)=\sum_{n\geq 1}\frac{1}{(n+z)^2}\)

...but there exists a more elementary and much simpler method, can you figure it out?

- Jan 17, 2013

- 1,667

I am sure there is but since I noticed your interest in integral and series, I wanted to give you another way of solving the question.Thanks ZaidAlyafey for your participation, your answer is correct.

...but there exists a more elementary and much simpler method, can you figure it out?

- Thread starter
- #5

- Nov 4, 2013

- 428

Thanks but currently, that kind of stuff is way above my level and I wonder if I will ever come across those Beta and Gamma functions.I am sure there is but since I noticed your interest in integral and series, I wanted to give you another way of solving the question.

- Feb 13, 2012

- 1,704

Compute:

$$\int_0^{\pi/2} \tan(x)\ln(\sin(x))\,dx$$

http://mathhelpboards.com/math-notes-49/integrals-natural-logarithm-5286.html

... You can use the identity...

$\displaystyle \int_{0}^{1} x^{n}\ \ln x\ d x = - \frac{1}{(n+1)^{2}}\ (1)$

With the substitution $\sin x = u$ and taking into account (1) the integral becomes...

$\displaystyle \int_0^{\pi/2} \tan x\ \ln \sin x\ d x = \int_{0}^{1} \frac{u}{1 - u^{2}}\ \ln u\ d u = \sum_{n = 0}^{\infty} \int_{0}^{1} u^{2 n + 1}\ \ln u\ d u = - \sum_{n = 0}^{\infty} \frac{1}{(2 n + 2)^{2}} = - \frac{\pi^{2}}{24}\ (2)$

Kind regards

$\chi$ $\sigma$

- Thread starter
- #7

- Nov 4, 2013

- 428

Great!

http://mathhelpboards.com/math-notes-49/integrals-natural-logarithm-5286.html

... You can use the identity...

$\displaystyle \int_{0}^{1} x^{n}\ \ln x\ d x = - \frac{1}{(n+1)^{2}}\ (1)$

With the substitution $\sin x = u$ and taking into account (1) the integral becomes...

$\displaystyle \int_0^{\pi/2} \tan x\ \ln \sin x\ d x = \int_{0}^{1} \frac{u}{1 - u^{2}}\ \ln u\ d u = \sum_{n = 0}^{\infty} \int_{0}^{1} u^{2 n + 1}\ \ln u\ d u = - \sum_{n = 0}^{\infty} \frac{1}{(2 n + 2)^{2}} = - \frac{\pi^{2}}{24}\ (2)$

Kind regards

$\chi$ $\sigma$

I took a slightly different approach:

Use the substitution $\cos x=u$ to get,

$$\int_0^{1} \frac{\ln(\sqrt{1-u^2})}{u}\,du=-\int_0^1 \frac{1}{2u}\left(\sum_{k=1}^{\infty} \frac{u^{2k}}{k}\right)\,du=-\sum_{k=1}^{\infty} \int_0^1 \frac{u^{2k-1}}{2k}\,du$$

$$=-\frac{1}{4}\sum_{k=1}^{\infty}\frac{1}{k^2}=-\frac{\zeta(2)}{4}$$