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Find $$\int _0^{\infty }\frac{x^{\frac{4}{5}}-x^{\frac{2}{3}}}{\ln \left(x\right)\left(x^2+1\right)}\:dx.$$

I'd like to know in what ways can one approach this integral that can be found here, since the post was about using feynman's trick to evaluate integrals i used the parameter, $$I=\int _0^{\infty }\frac{x^{\frac{4}{5}}-x^{\frac{2}{3}}}{\ln \left(x\right)\left(x^2+1\right)}\:dx$$ $$I\left(a\right)=\int _0^{\infty }\frac{x^{\frac{4}{5}a}-x^{\frac{2}{3}}}{\ln \left(x\right)\left(x^2+1\right)}\:dx$$ $$I'\left(a\right)=\frac{4}{5}\int _0^{\infty }\frac{x^{\frac{4}{5}a}}{x^2+1}\:dx$$

where $I\left(a=1\right)=I$ and $I\left(a=\frac{5}{6}\right)=0$.

But that integral doesnt seem so simple to tackle. i'd appreciate any ideas or different approaches to the integral.

Batominovski
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Dennis Orton
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3 Answers3

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Consider $$J(a)=\int_0^\infty\frac{x^a-1}{(x^2+1)\ln x}dx$$ where $a\in\Bbb C$ with $|\Re a|<1$. Then $$J'(a)=\int_0^\infty\frac{x^a}{x^2+1}dx.$$ Using the keyhole contour you can see that $$J'(a)=\frac{2\pi i}{1-e^{2\pi a}}\left(\frac{e^{\frac{\pi a}{2}}}{2i}+\frac{e^{\frac{3\pi a}{2}}}{-2i}\right)=\frac{\pi}{2}\sec\frac{\pi a}{2}.$$

Alternatively with $y=x^2$ and $u=\frac{y}{y+1}$, note that \begin{align}J'(a)&=\frac{1}{2}\int_0^\infty \frac{y^{\frac{a+1}{2}-1}}{y+1}dy=\frac12\int_0^1 u^{\frac{a+1}{2}-1}(1-u)^{\left(1-\frac{a+1}{2}\right)-1}du\\&=\frac12\mathrm{B}\left(\frac{a+1}{2},1-\frac{a+1}{2}\right)=\frac12\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(1-\frac{a+1}{2}\right).\end{align} From Euler's reflection formula, $J'(a)=\frac{\pi}{2}\operatorname{cosec}\left(\pi\frac{a+1}{2}\right)=\frac{\pi}{2}\sec\frac{\pi a}{2}$.

Therefore $$J(a)=\int_0^a \frac{\pi}{2}\sec \frac{\pi t}{2} dt=\int_0^{\frac{\pi a}{2}} \sec\theta d\theta=\ln\left(\tan\frac{\pi a}{2}+\sec\frac{\pi a}{2}\right).$$ Therefore $$I=J\left(\frac45\right)-J\left(\frac23\right)=\ln\frac{\tan\frac{2\pi}{5}+\sec \frac{2\pi}{5}}{\tan\frac{\pi}{3}+\sec{\frac{\pi}{3}}}=\ln\frac{\sqrt{5+2\sqrt{5}}+1+\sqrt{5}}{\sqrt3 +2}\approx 0.525772.$$

Batominovski
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  • Very nice approach with the beta function, i always thought it was somehow involved in the integral, i ended up doing almost exactly the same but instead i used the definition $\mathrm{B}\left(t,z\right)=\int _0^{\infty }\frac{x^{t-1}}{\left(x+1\right)^{t+z}}:dx$. – Dennis Orton Apr 06 '20 at 23:06
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$$I'\left(a\right)=\frac{4}{5}\int _0^{\infty }\frac{x^{\frac{4}{5}a}}{x^2+1}\,dx=\frac{2\pi}{5} \sec \left(\frac{2 \pi }{5}a\right)\qquad \text{if} \quad -\frac{5}{4}<\Re(a)<\frac{5}{4}$$

Use the tangent half-angle substitution for $I(a)$.

Claude Leibovici
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Hints. You may evaluate $I'(a)$ by using the substitution $x=\sinh\phi,$ which reduces the integral to a constant multiple of the form $$\int_0^{+\infty}\sinh^{4a/5}\phi\mathrm d\phi.$$ This may be rewritten as $$\int_0^{+\infty}\left(\frac{e^{\phi}}{2}\right)^{4a/5}(1-e^{-2\phi})^{4a/5}\phi\mathrm d\phi,$$ which you may expand by the binomial theorem. It converges for all $\phi>0.$

Allawonder
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