TITLE: Question related to beta and gamma function
QUESTION [2 upvotes]: I'm trying to derive the following integral.
$$\int_0^\infty \frac{x^8(1-x^6)}{(1+x)^{24}} \, dx.$$
What transformations can I use?

REPLY [0 votes]: Here is a way to evaluate the integral without invoking the Beta function.
Writing the integral as
$$\int_0^\infty \frac{x^8 (1 - x^6)}{(1 + x)^{24}} \, dx = \int_0^1 \frac{x^8 (1 - x^6)}{(1 + x)^{24}} \, dx + \int_1^\infty \frac{x^8 (1 - x^6)}{(1 + x)^{24}} \, dx.$$
In the second of these integral, if a substitution of $x \mapsto 1/x$ is enforced we have
\begin{align*}
\int_0^\infty \frac{x^8 (1 - x^6)}{(1 + x)^{24}} \, dx &= \int_0^1 \frac{x^8 (1 - x^6)}{(1 + x)^{24}} \, dx + \int_0^1 \frac{\frac{1}{x^8} \left (1 - \frac{1}{x^6} \right )}{\left (1 + \frac{1}{x} \right )^{24}} \frac{dx}{x^2}\\
&= \int_0^1 \frac{x^8 (1 - x^6)}{(1 + x)^{24}} \, dx + \int_0^1 \frac{x^8 (x^6 - 1)}{(1 + x)^{24}} \, dx\\
&= \int_0^1 \frac{x^8 (1 - x^6)}{(1 + x)^{24}} \, dx - \int_0^1 \frac{x^8 (1 - x^6)}{(1 + x)^{24}} \, dx\\
&= 0.
\end{align*}
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