Integration of an infinite sum How do you calculate:  $$\int_0^1 \left(\sum_{n=0}^\infty x^{2 n + 1} t^n (1 - t)^n\right) dt \text{?}$$
We can write
$$I = \sum_{n=0}^\infty x^{2n+1}\int_0^1 t^n(1-t)^ndt$$
also by definition of beta function
$$B(a+1,b+1) = \int_0^1t^a(1-t)^bdt = \frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+2)}$$
here $a=b=n$ and $n$ is an integer
this gives
$$I = \sum_{n=0}^\infty x^{2n+1}\frac{\Gamma(n+1)^2}{\Gamma(2n+2)}$$
$$I = \sum_{n=0}^\infty x^{2n+1}\frac{(n!)^2}{(2n+1)!}$$
By ratio test
$$\lim_{n\to\infty}\left| \frac{a_{n+1}}{a_n}\right|= r$$
it comes out that series converges for $x&lt;2$.
The sum converges for $|x^2t(1-t)|&lt;1$. After that, you are left with the integrand that looks like
$$
\frac{1}{1-x^2t+x^2t^2}
$$
Close the square to get
$$
\frac{1}{(t-0.5)^2 - 0.25+\frac{1}{x^2}}
$$
which is an integral of the form (after substituting $v=t-0.5$
$$
\int_{0}^{1}\frac{dv}{v^2-a^2}
$$