Investigate on the convergence of $I(a)= \int_0^{\infty} \frac{x}{\sqrt{1+x^a}} d x $ for $a>0$ and its exact value in case of convergence.

Question

Inspired by the post, I start to investigate the convergence of $$I(a)= \int_0^{\infty} \frac{x}{\sqrt{1+x^a}} d x $$ for $a>0$.

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For any $0\le a\le 4$, if $x\ge 1$, we have

$$ \frac{x}{\sqrt{1+x^a}}\ge\frac{x}{\sqrt{1+x^4}}=\frac{1}{\sqrt{x^2+\frac{1}{x^2}}}\ge \frac{1}{\sqrt{x^2+x^2}} =\frac{1}{x\sqrt2} $$

Since $\int_0^{\infty} \frac{1}{x \sqrt{2}} d x$ is divergent, therefore $ I(a)$ is divergent for any $0\le a\le 4$.

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For $a>4$, let $x=\tan ^{\frac{2}{a}} \theta$, then $$ \begin{aligned} I(a)& =\frac{2}{a} \int_0^{\frac{\pi}{2}} \frac{\tan ^{\frac{2}{a}} \theta}{\sec \theta} \cdot \tan ^{\frac{2}{a}-1} \theta \sec ^2 \theta d \theta \\ & =\frac{2}{a} \int_0^{\frac{\pi}{2}} \tan ^{\frac{4}{a}-1} \theta \sec \theta d \theta \\ & =\frac{2}{n} \int_0^{\frac{\pi}{2}} \sin ^{\frac{4}{a}-1} \theta \cos ^{-\frac{4}{a}} \theta d \theta \\ & =\frac{1}{a} B\left(\frac{2}{a}, \frac{1}{2}-\frac{2}{a}\right) \\ &=\frac{\Gamma\left(\frac{2}{a}\right) \Gamma\left(\frac{1}{2}-\frac{2}{a}\right)}{a \sqrt{\pi}} \end{aligned} $$ which are convergent $ \Leftrightarrow \frac{1}{2}-\frac{2}{a}>0 \Leftrightarrow a>4$.

Comments and alternative solutions are highly appreciated!

Answer 1

The antiderivative $$\int \frac{x}{\sqrt{1+x^a}} dx=\frac{1}{2} x^2 \, _2F_1\left(\frac{1}{2},\frac{2}{a};\frac{a+2}{a};-x ^a\right)$$ leads to the same result for the definite integral.

What is interesting to notice is that, if $a=4+\epsilon$, the definite integral can be approximated as $$\int_0^\infty \frac{x}{\sqrt{1+x^{4+\epsilon}}} dx=\frac{2}{\epsilon }+\frac{\log (2)}{2}+\frac{\pi ^2-6 (2-\log (2)) \log (2)}{96} \epsilon +O\left(\epsilon ^2\right)$$ Making $\epsilon=1$ gives $$\frac{\pi ^2}{96}+\frac{32+6 \log (2)+\log ^2(2)}{16}= 2.39277$$ while $$\frac{\Gamma \left(\frac{1}{10}\right) \Gamma \left(\frac{7}{5}\right)}{2 \sqrt{\pi }}=2.38116$$

Answer 2

Similarly to the question you asked in, the Mellin transform of $f(x)=(1+x)^{-\rho}$ is

$$ \tilde{f}(s) = \mathcal{M}[f(x);s] = \frac{1}{\Gamma(\rho)}\Gamma(s)\Gamma(\rho-s) $$

which is defined for $0<\Re(s)<\Re(\rho)$. Then, by the Mellin transform, one has $f(x^n) \rightarrow \frac{1}{|n|}\tilde{f}\left(\frac{s}{n} \right)$ that

$$ \int_0^{\infty} \frac{x^{s-1}}{(1+x^n)^{\rho}}dx = \frac{\Gamma\left(\frac{s}{n}\right)\Gamma\left(\rho-\frac{s}{n}\right)}{n\Gamma(\rho)} $$

which is defined for $0<\Re(\frac{s}{n})<\Re(p)$, and in the specific case above gives that (with $s=2,n=a,\rho=1/2)$

$$ \int_0^{\infty} \frac{x}{\sqrt{1+x^a}} dx = \frac{\Gamma(\frac{2}{a})\Gamma(\frac{1}{2}-\frac{2}{a})} {a\sqrt{\pi}} $$

which is defined for $0<\Re(\frac{2}{a})<\frac{1}{2}$. i.e. $a>4$ as you noted.