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Post Mon Sep 12, 2016 7:02 pm

Posts: 82
If \(\displaystyle \displaystyle I = \int^{\pi}_{0}x^3\ln(\sin x)dx\) and \(\displaystyle \displaystyle J = \int^{\pi}_{0}x^2\ln(\sqrt{2}\sin x)dx\;,\)Then \(\displaystyle \displaystyle \frac{I}{J} =\)

Post Thu Oct 06, 2016 3:00 pm
FDP

Posts: 22
For $n\in \mathbb{N}$, let,

$\displaystyle A_n=\int_0^{\pi} x^n\ln(\sin x)dx$

$\displaystyle A_0=\int_0^{\pi} \ln(\sin x)dx=\int_0^{\tfrac{\pi}{2}} \ln(\sin x)dx+ \int_{\tfrac{\pi}{2}}^{\pi} \ln(\sin x)dx$

In the latter integral perform the change of variable $y=\pi-x$,

$\displaystyle A_0=2\int_0^{\tfrac{\pi}{2}} \ln(\sin x)dx$

Perform the change of variable $y=\cot x$,

$\displaystyle A_0=-\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}dx$

Consider the function $F$ defined by,

for $x\in[0;1]$,

$\displaystyle F(y)=\int_0^1 \dfrac{\ln(1+x^2y^2)}{1+x^2}dx$

Observe that $F(0)=0$ and $A_0=-F(0)$


$\displaystyle F(1)=F(1)-F(0)\\$
$\displaystyle F(1)=\int_0^1 F^\prime(y)dy\\$
$\displaystyle F(1)=\int_0^1 \left(\int_0^{+\infty}\dfrac{2yx^2}{(1+x^2)(1+x^2y^2)}dx\right)dy\\$
$\displaystyle F(1)=\int_0^1 \left(\left[2y\left(\dfrac{\arctan x}{y^2-1}-\dfrac{\arctan(xy)}{y(y^2-1)}\right)\right]_{x=0}^{x=+\infty}\right)dy$
$\displaystyle F(1)=\pi\int_0^1 \dfrac{1}{1+y}dy$
$\displaystyle F(1)=\pi\ln 2$

Therefore,
$\boxed{A_0=-\pi\ln 2}$

Consider $n=2$ and perform the change of variable $y=\pi-x$,

$\displaystyle A_2=\int_0^{\pi} (\pi-x)^2\ln(\sin x)dx=\pi^2A_0-2\pi A_1+A_2$

Therefore,

$\boxed{A_1=\dfrac{\pi}{2}A_0=-\dfrac{\pi^2}{2}\ln 2}$

Consider $n=3$, performing the same change of variable leads to:

$A_3=\pi^3A_0-3\pi^2A_1+3\pi A_2-A_3$

Therefore,

$\boxed{A_3=-\dfrac{\pi^3}{4}A_0+\dfrac{3\pi}{2}A_2=\dfrac{3\pi}{2}\left(A_2+\dfrac{\pi^3}{6} \ln 2\right)}$

On the other hand,

$\boxed{I=A_3}$

$\displaystyle J=A_2+\int_0^\pi x^2\ln(\sqrt{2})dx$

Therefore,
$\boxed{J=A_2+\dfrac{\pi^3}{6} \ln 2}$

and,

$\boxed{\dfrac{I}{J}=\dfrac{3\pi}{2}}$

Post Tue Nov 22, 2016 5:27 pm

Posts: 82
Thanks FDP for nice solution.


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