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Evaluation of some Definite Integrals

(a) \(\displaystyle \int_{0}^{\infty}\frac{x^3+3}{x^6\cdot (x^2+1)}dx\)

(b) \(\displaystyle \int_{0}^{\infty}\frac{x\cdot e^{-x}}{\sqrt{1-e^{-2x}}}dx\)

\((b)\) \begin{align*}
\displaystyle\int_{0}^{\infty}{\frac{x\,{\rm{e}}^{-x}}{\sqrt{1-{\rm{e}}^{-2x}}}dx}&\mathop{=\!=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{\sin{t}\,=\,{\rm{e}}^{-x}}\\
{-\frac{\cos{t}}{\sin{t}} dt\,=\,dx}\\
\end{subarray}}\,-\int_{0}^{\frac{\pi}{2}}{\frac{\log({\sin{t}})\,\sin{t}}{\cos{t}}\,\frac{\cos{t}}{\sin{t}} dt}\\
&=-\int_{0}^{\frac{\pi}{2}}{\log({\sin{t}})\, dt}\\
&\stackrel{(*)}{=}\frac{\pi}{2}\log2\,.
\end{align*}

\((*)\) Easy to prove. Left as an exercise.

Thanking You Grigorios Kostakos Nice Solution.

Le me prove this Grigoris, \( \displaystyle -\int_{0}^{\pi/2}\ln \left ( \sin x \right )\, dx\) just in case it appears again, so that we give it as a link.

Applying the sub \( \displaystyle u=\frac{\pi}{2}-x \) gives us:

\( \begin{aligned}
\int_{0}^{\pi/2}\ln \left ( \sin x \right )\, dx &\overset{u=\pi/2-x}{=\! =\! =\! =\! =\! } \int_{0}^{\pi/2}\ln \left ( \sin \left ( \frac{\pi}{2}-u \right ) \right ), du\\
&= \int_{0}^{\pi/2}\ln \left ( \cos u \right )\, du\\
&= \int_{0}^{\pi/2}\ln \left ( \cos x \right )\, dx\\
\end{aligned} \)

Adding by parts we get:
$$\left.\begin{matrix}
\displaystyle J=\int_{0}^{\pi/2}\ln \left ( \sin x \right )\, dx & \\
\displaystyle J=\int_{0}^{\pi/2}\ln \left ( \cos x \right )\, dx&
\end{matrix}\right\}(+)\Rightarrow 2J=\int_{0}^{\pi/2} \ln \left ( \sin x\cos x \right )\, dx\Leftrightarrow 2J=\int_{0}^{\pi/2}\ln \left ( \frac{\sin 2x}{2} \right )\, dx$$
$$\begin{aligned}
2J=\int_{0}^{\pi/2}\ln \left ( \frac{\sin 2x}{2} \right )\, dx &\iff 2J=\int_{0}^{\pi/2}\ln \left ( \sin 2x \right )\, dx-\int_{0}^{\pi/2}\ln 2\, dx \\
&\iff 2J=\int_{0}^{\pi/2}\ln \left ( \sin 2x \right )\, dx-\frac{\pi\ln 2}{2} \\
&\overset{u=2x}{\iff}2J=\frac{1}{2}\int_{0}^{\pi}\ln \left ( \sin x \right )\, dx-\frac{\pi\ln 2}{2}\\
&\iff 2J= \int_{0}^{\pi/2}\ln \left ( \sin x \right )-\frac{\pi\ln 2}{2} \\
&\iff 2J=J-\frac{\pi\ln 2}{2} \\
&\iff J=-\frac{\pi\ln 2}{2}
\end{aligned}$$

Thus \( \displaystyle -\int_{0}^{\pi/2}\ln \left ( \sin x \right )\, dx=\frac{\pi\ln 2}{2} \).

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For reasons of variety let me write another proof using \( {\rm B}, \Gamma \) functions.

First of all we apply the sub \( u=\sin^2 x\) thus we get:
$$\int_{0}^{\pi/2}\ln \left ( \sin x \right )\, dx\overset{u=\sin x}{=\! =\! =\! =\!}\frac{1}{4}\int_{0}^{1}\frac{\ln u}{\sqrt{u\left ( 1-u \right )}}\, du\overset{u={\rm 1-v}}{=\! =\! =\! =\!}\frac{1}{4}\int_{0}^{1}\frac{\ln \left ( 1-{\rm v} \right )}{\sqrt{{\rm v\left ( 1-v \right )}}}\, d{\rm v}$$

Adding the two last integrals and dividng by two we get that:
$$\int_{0}^{\pi/2}\ln \left ( \sin x \right )\, dx=\frac{1}{4}\int_{0}^{1}\frac{\ln \sqrt{u\left ( 1-u \right )}}{\sqrt{u\left ( 1-u \right )}}\, du=-\frac{1}{4}\frac{\mathrm{d} }{\mathrm{d} s}{\rm B}\left ( s, s \right )\bigg|_{s=\frac{1}{2}}$$

It is known that: \( \displaystyle {\rm B}\left ( s, s \right )=\frac{\Gamma (s)\Gamma (s)}{\Gamma (s+s)}=\frac{\Gamma^2(s)}{\Gamma (2s)} \) and \( \displaystyle \frac{\mathrm{d} }{\mathrm{d} s}{\rm B}(s, s)=2{\rm B}(s, s)\left ( \psi (s)-\psi (2s) \right ) \). Also it holds that \( \displaystyle {\rm B}\left ( \frac{1}{2}, \frac{1}{2} \right )=\frac{\Gamma \left ( \frac{1}{2} \right )\Gamma \left ( \frac{1}{2} \right )}{\Gamma \left ( 1 \right )}=\frac{\sqrt{\pi}\sqrt{\pi}}{1}=\pi \) and \( \displaystyle 2\psi (2s)=\psi (s)+\psi \left ( s+\frac{1}{2} \right )+2\ln 2 \).The last equation for \( s=1/2 \) gives: \( \psi (1/2)-\psi (1)=-2\ln 2 \).

Combining them all together gives that: \( \displaystyle \int_{0}^{\pi/4}\ln \left ( \sin x \right )\, dx=-\frac{\pi\ln 2}{2} \).

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There is also a solution using complex analysis.
However I don't see the point of writting that, since it is a too long solution, and second because we already have two.

Thanking You Apostolos J. Kos, Yours (II) Method is very Nice.

**My Solution::** Given \(\displaystyle \int_{0}^{\frac{\pi}{2}}\ln (\sin x)dx\)

Now Let \(\displaystyle \sin x = \left(\frac{e^{ix}-e^{-ix}}{2i}\right) = \frac{1}{2i}\cdot \left(\frac{e^{2ix}-1}{e^{ix}}\right)\)

So Integral Convert into \(\displaystyle \int_{0}^{\frac{\pi}{2}}\ln \left(\frac{e^{2ix}-1}{2i\cdot e^{ix}}\right)dx = \int_{0}^{\frac{\pi}{2}}\ln(e^{2ix}-1)dx-\int_{0}^{\frac{\pi}{2}}\ln(e^{ix})dx-\int_{0}^{\frac{\pi}{2}}\ln(2)dx-\int_{0}^{\frac{\pi}{2}}\ln(i)dx\)

\(\displaystyle = \int_{0}^{\frac{\pi}{2}}\ln(e^{2ix}-1)dx-i\cdot \frac{\pi^2}{8}-\frac{\pi}{2}\cdot \ln(2)-i\cdot \frac{\pi^2}{4}\)

Now we will Solve \(\displaystyle I = \int_{0}^{\frac{\pi}{2}}\ln(e^{2ix}-1)dx\)

Using \(\displaystyle e^{2ix}-1=\cos(2x)+i\sin(2x)-1 = i\cdot 2\sin x\cdot \cos x-2\sin^2 (x) = 2i\sin x\cdot (\cos x+i\sin x) = 2i\sin x\cdot e^{ix}\)

So \(\displaystyle I = \int_{0}^{\frac{\pi}{2}}\ln(2i\cdot \sin x)dx+\int_{0}^{\frac{\pi}{2}}\ln(e^{ix})dx\)

So \(\displaystyle I = \int_{0}^{\frac{\pi}{2}}\ln(2i\cdot \sin x)dx+i\cdot \frac{\pi^2}{8}\)

So \(\displaystyle \int_{0}^{\frac{\pi}{2}}\ln(\sin x)dx = -\frac{\pi}{2}\cdot \ln(2)+\bf{Imaginary\; quantity.}\)

But we have calculate only for real values.

So \(\displaystyle \int_{0}^{\frac{\pi}{2}}\ln(\sin x)dx = -\frac{\pi}{2}\cdot \ln(2)\)

**My Try for \(\displaystyle (a)\;\; I = \int_{0}^{\infty}\frac{x^3+3}{x^6\cdot \left(x^2+1\right)}dx\)

Now Let \(\displaystyle x = \frac{1}{t}\) and \(\displaystyle dx = -\frac{1}{t^2}\) and Changing Limit, We Get

\(\displaystyle I = \int_{0}^{\infty}\frac{(1+3t^3)\cdot t^6\cdot t^2}{t^3\cdot (1+t^2)}\cdot -\frac{1}{t^2}dt = \int_{0}^{\infty}\frac{t^3+3t^6}{1+t^2}dt\)

Now How can I solve after that

Help me

Thanks

\((a)\) Jacks the integral \(\displaystyle \int_{0}^{\infty}{\frac{x^3+3}{x^6(x^2+1)}dx}\) does not converges. And here is why:
\begin{align*}
I&=\displaystyle \int_{0}^{\infty}{\frac{x^3+3}{x^6(x^2+1)}dx}\\
&=\int_{0}^{1}{\frac{x^3+3}{x^6(x^2+1)}dx}+\int_{1}^{\infty}{\frac{x^3+3}{x^6(x^2+1)}dx}\\
&=I_1+I_2\,.
\end{align*}
Because \((\forall x\in(1,+\infty))\displaystyle\quad0<\frac{x^3+3}{x^6(x^2+1)}<\frac{x^3+3}{x^6}\) and \[\displaystyle\int_{1}^{\infty}{\frac{x^3+3}{x^6}dx}=\frac{11}{10}\] the \(I_1\) converges, i.e. is a real number.

Because \((\forall x\in(0,1))\quad\displaystyle\frac{x^3+3}{x^6(x^2+1)}>\frac{x^3+3}{x^6(x^3+3)}=\frac{1}{x^6}\) and \[\displaystyle\int_{0}^{1}{\frac{1}{x^6}dx}=+\infty\] the \(I_2\) diverges.

So \(I=I_1+I_2\) diverges.

Thanking You Grigorios Kostakos , Got it.