Improper integral $\int_0^\infty \frac{x^{\alpha}\ln x}{x^2+1}\,dx=\frac{\pi^2}{4} \frac{\sin(\pi \alpha/2)}{\cos^2(\pi \alpha/2)}$

Question

$$\int_0^\infty \frac{x^{\alpha}\ln x}{x^2+1}\,dx=\frac{\pi^2}{4} \frac{\sin(\pi \alpha/2)}{\cos^2(\pi \alpha/2)}$$ where $0 < \alpha < 1$.

Answer: When i put this term in my integral calculator, it gave me very lengthy answer involving polylogarithm functions.

@Nosrati, I didn't search in this site about this integral. But if this is a duplicate question, please provide the link of this question's answer available in this site. — Win_odd Dhamnekar, Nov 19 '18 at 04:08

score 6 · Accepted Answer · answered Nov 19 '18 at 07:25

$$ \begin{align} \int_0^\infty\frac{x^\alpha\log(x)}{x^2+1}\,\mathrm{d}x &=\frac{\mathrm{d}}{\mathrm{d}\alpha}\int_0^\infty\frac{x^\alpha}{x^2+1}\,\mathrm{d}x\tag1\\ &=\frac12\frac{\mathrm{d}}{\mathrm{d}\alpha}\int_0^\infty\frac{x^{\frac{\alpha-1}2}}{x+1}\,\mathrm{d}x\tag2\\ &=\frac12\frac{\mathrm{d}}{\mathrm{d}\alpha}\frac{\Gamma\left(\frac{1+\alpha}2\right)\Gamma\left(\frac{1-\alpha}2\right)}{\Gamma(1)}\tag3\\ &=\frac\pi2\frac{\mathrm{d}}{\mathrm{d}\alpha}\csc\left(\pi\frac{1-\alpha}2\right)\tag4\\ &=\frac\pi2\frac{\mathrm{d}}{\mathrm{d}\alpha}\sec\left(\frac{\pi\alpha}2\right)\tag5\\ &=\frac{\pi^2}4\tan\left(\frac{\pi\alpha}2\right)\sec\left(\frac{\pi\alpha}2\right)\tag6 \end{align} $$ Explanation:
$(1)$: $\frac{\mathrm{d}}{\mathrm{d}\alpha}x^\alpha=x^\alpha\log(x)$
$(2)$: substitute $x\mapsto\sqrt{x}$
$(3)$: Beta Function
$(4)$: Euler Reflection Formula
$(5)$: trigonometric identity
$(6)$: evaluate the derivative

How did you compute (2)? You substitute x by $\sqrt{x}$.So I get $\frac12\frac{d}{d\alpha}\int\limits_0^\infty\frac{x^\frac{\alpha}{2}}{x+1}dx$ But your term in (2) is different? — Win_odd Dhamnekar, Nov 19 '18 at 13:28
What is $\mathrm{d}\sqrt{x}$? You got the $\frac12$, but forgot the $x^{-1/2}$. — robjohn, Nov 19 '18 at 13:58
I already got the answers to this question from you and one other member of this site correctly satisfying all my queries. Then what is the use of making it off-topic and put on hold? — Win_odd Dhamnekar, Nov 25 '18 at 12:35

Nosrati · Answer 2 · 2018-11-19T05:33:31.037

4

By substitution $x=\tan t$ $$I(a)=\int_0^\infty\dfrac{x^a}{1+x^2}\ dx=\int_0^{\pi/2}\tan^at\ dt$$ then using dear Beta function $$I(a)=\dfrac12\Gamma\left(\dfrac{a+1}{2}\right)\Gamma\left(\dfrac{-a+1}{2}\right)=\dfrac{\pi}{2\sin\pi\dfrac{1-a}{2}}=\dfrac{\pi}{2\cos\dfrac{a\pi}{2}}$$ by Reflection formula the desired integral is $$\dfrac{d}{da}I(a)=\color{blue}{\dfrac{\pi^2}{4}\dfrac{\sin\frac{a\pi}{2}}{\cos^2\frac{a\pi}{2}}}$$

edited Nov 19 '18 at 05:33

answered Nov 19 '18 at 04:15

Nosrati

29,995

1

How did you compute the first equation? I never found it in the list of trigonometric identities? – Win_odd Dhamnekar Nov 19 '18 at 04:59
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It is in fact no trigonometric identity but just the substitution $x=\tan(t)$. From hereon you will get the upper as well as the lower border of integration. Further note that the derivative of $x=\tan(t)$ can be written as $dt=\frac{dx}{1+x^2}$ due the derivative of the tangent function. – mrtaurho Nov 19 '18 at 10:48
@Nosrati, is $I(a)$ regularised incomplete beta function? why it is multiplied by$\frac12$? – Win_odd Dhamnekar Nov 20 '18 at 05:35
See the link, fourth formula of Properties. – Nosrati Nov 20 '18 at 05:37
@Nosrati, But it is $I_x(1,b)=1-(1-x)^b$. Now what is x and what is b? I assume that it is a derivative of $\tan^{\frac{a-1}{2}}{(x)}$ – Win_odd Dhamnekar Nov 20 '18 at 06:11
Is $\int_0^\infty\dfrac{x^a}{1+x^2}\ dx=\int_0^{\pi/2}\tan^at\ dt$ clear? – Nosrati Nov 20 '18 at 06:16
@Nosrati, Yes, You have substituted x=tan(t). – Win_odd Dhamnekar Nov 20 '18 at 06:31
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So $$I(a)=\int_0^{\pi/2}\tan^at\ dt=\dfrac12\left(2\int_0^{\pi/2}\sin^at\cos^{-a}t\ dt\right)$$ – Nosrati Nov 20 '18 at 06:36
@Nosrati, That's the very good math trick. – Win_odd Dhamnekar Nov 20 '18 at 07:01

Improper integral $\int_0^\infty \frac{x^{\alpha}\ln x}{x^2+1}\,dx=\frac{\pi^2}{4} \frac{\sin(\pi \alpha/2)}{\cos^2(\pi \alpha/2)}$

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