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.

Answer 1

$$ \begin{align} \int_0^\infty\frac{x^\alpha\log(x)}{x^2+1}\,\mathrm{d}x &=\frac{\mathrm{d}}{\mathrm{d}\alpha}\int_0^\infty\frac{x^\alpha}{x^2+1}\,\mathrm{d}x\tag1\\ &=\frac12\frac{\mathrm{d}}{\mathrm{d}\alpha}\int_0^\infty\frac{x^{\frac{\alpha-1}2}}{x+1}\,\mathrm{d}x\tag2\\ &=\frac12\frac{\mathrm{d}}{\mathrm{d}\alpha}\frac{\Gamma\left(\frac{1+\alpha}2\right)\Gamma\left(\frac{1-\alpha}2\right)}{\Gamma(1)}\tag3\\ &=\frac\pi2\frac{\mathrm{d}}{\mathrm{d}\alpha}\csc\left(\pi\frac{1-\alpha}2\right)\tag4\\ &=\frac\pi2\frac{\mathrm{d}}{\mathrm{d}\alpha}\sec\left(\frac{\pi\alpha}2\right)\tag5\\ &=\frac{\pi^2}4\tan\left(\frac{\pi\alpha}2\right)\sec\left(\frac{\pi\alpha}2\right)\tag6 \end{align} $$ 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

Answer 2

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 $$\dfrac{d}{da}I(a)=\color{blue}{\dfrac{\pi^2}{4}\dfrac{\sin\frac{a\pi}{2}}{\cos^2\frac{a\pi}{2}}}$$