$\int_{0}^{\infty}{\frac{x^{a}}{1+x^{a}}dx}$

Calculus (Integrals, Series)
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Grigorios Kostakos
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$\int_{0}^{\infty}{\frac{x^{a}}{1+x^{a}}dx}$

#1

Post by Grigorios Kostakos »

For the values of \(a\) for which the integral converges, calculate the integral \[\displaystyle\int_{0}^{+\infty}{\frac{x^{a}}{1+x^{a}}dx}\,.\]
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Tolaso J Kos
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Re: $\int_{0}^{\infty}{\frac{x^{a}}{1+x^{a}}dx}$

#2

Post by Tolaso J Kos »

Good evening Grigoris.

We apply the sub $x=y^{1/a}$ , thus $dx=\frac{1}{a}y^{1/a-1}dy$. So, if $a<0: x=0\Rightarrow y=\infty , \quad x=\infty \Rightarrow x=0$ , else if $a>0: x=0\Rightarrow y=0, \quad x=\infty \Rightarrow y=\infty$. Therefore:

$$\begin{align*}
\int_{0}^{\infty} \frac{x^a}{1+x^a}\, {\rm d}x&\; \overset{x=y^{1/a}}{=\! =\! =\! =\! =\!} \; \frac{1}{\left | a \right |}\int_{0}^{\infty}\frac{y^{1/a}}{1+y}\, {\rm d}y \\
&=\frac{1}{\left | a \right |}\int_{0}^{\infty}\frac{y^{p-1}}{\left ( 1+y \right )^{p+q}}\, {\rm d}y \quad \quad \quad p=1/a +1, \;\; q= -1/a \\
&= \frac{1}{|a|}{\rm B}\left ( p, q \right )\\
&\overset{a<-1}{=\! =\! =\!}- \frac{1}{a}\frac{\Gamma \left ( \frac{1}{a}+1 \right )\Gamma \left ( -\frac{1}{a} \right )}{\Gamma(1)} \\
&= - \frac{1}{a}\Gamma \left ( - \frac{1}{a} \right ) \Gamma \left ( \frac{1}{a}+1 \right ) \\
&= \frac{\pi \csc \pi a}{a}
\end{align*}$$
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