MHB Mathematical Techniques for Solving the Definite Integral Challenge

AI Thread Summary
The integral $$\int_0^{\pi/2} \tan(x)\ln(\sin(x))\,dx$$ evaluates to $$-\frac{\pi^2}{24}$$ using differentiation under the integral sign and properties of the Gamma function. The discussion highlights two methods to arrive at this result, one involving the Gamma function and the other utilizing a substitution and series expansion. Both approaches confirm the same final value. Participants express interest in exploring simpler methods for solving the integral. The conversation emphasizes the mathematical techniques and identities used in the evaluation process.
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Compute:
$$\int_0^{\pi/2} \tan(x)\ln(\sin(x))\,dx$$
 
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[sp]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 $$x=1$$ 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 $$\Gamma (1+z)= z\Gamma(z)$$

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

[/sp]
 
ZaidAlyafey said:
[sp]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 $$x=1$$ 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 $$\Gamma (1+z)= z\Gamma(z)$$

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

[/sp]

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

...but there exists a more elementary and much simpler method, can you figure it out? ;)
 
Pranav said:
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.
 
ZaidAlyafey said:
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. :o
 
Pranav said:
Compute:
$$\int_0^{\pi/2} \tan(x)\ln(\sin(x))\,dx$$

[sp]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...

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

Kind regards

$\chi$ $\sigma$
 
chisigma said:
[sp]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...

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

Kind regards

$\chi$ $\sigma$

Great! (Yes)

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