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Definite Integral challenge

Pranav

Well-known member
Nov 4, 2013
428
Compute:
$$\int_0^{\pi/2} \tan(x)\ln(\sin(x))\,dx$$
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Start by

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

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

 

Pranav

Well-known member
Nov 4, 2013
428
Start by

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

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

Thanks ZaidAlyafey for your participation, your answer is correct. :)

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

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Thanks ZaidAlyafey for your participation, your answer is correct. :)

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

Pranav

Well-known member
Nov 4, 2013
428
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 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. :eek:
 

chisigma

Well-known member
Feb 13, 2012
1,704
Compute:
$$\int_0^{\pi/2} \tan(x)\ln(\sin(x))\,dx$$
Proceeding as in...

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

$\chi$ $\sigma$
 

Pranav

Well-known member
Nov 4, 2013
428
Proceeding as in...

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

$\chi$ $\sigma$
Great! (Yes)

I took a slightly different approach:

Use the substitution $\cos x=u$ to get,
$$\int_0^{1} \frac{\ln(\sqrt{1-u^2})}{u}\,du=-\int_0^1 \frac{1}{2u}\left(\sum_{k=1}^{\infty} \frac{u^{2k}}{k}\right)\,du=-\sum_{k=1}^{\infty} \int_0^1 \frac{u^{2k-1}}{2k}\,du$$
$$=-\frac{1}{4}\sum_{k=1}^{\infty}\frac{1}{k^2}=-\frac{\zeta(2)}{4}$$