How to calculate indefinite integral involving infinite sums? I want to calculate the following integral:
$$ \int_{0}^{\infty}\left(x-\frac{x^3}{2}+\frac{x^5}{2\cdot 4}-\frac{x^7}{2\cdot 4\cdot 6}+\cdots\right)\;\left(1+\frac{x^2}{2^2}+\frac{x^4}{2^2\cdot 4^2}+\frac{x^6}{2^2\cdot 4^2\cdot 6^2}+\cdots\right)\,\mathrm{d}x 
$$
I have no idea how to start; any help is highly appreciated.
 A: A way is to consider the closed form of the series. Then, the integral of the product of them is a known Laplace transform :

A: Thank you, it was quite helpful; I know found an easier way to prove it:
Starting from $$\sum_{k=0}^{\infty} \frac{x^{2k+1}(-1)^k}{2^kk!}=xe^{-\frac{x^2}{2}}$$ we can rewrite the integral as follows:
$$ I=\int_{0}^{\infty}\left(x-\frac{x^3}{2}+\frac{x^5}{2\cdot 4}-\frac{x^7}{2\cdot 4\cdot 6}+\cdots\right)\;\left(1+\frac{x^2}{2^2}+\frac{x^4}{2^2\cdot 4^2}+\frac{x^6}{2^2\cdot 4^2\cdot 6^2}+\cdots\right)\,\mathrm{d}x = \int_{0}^{\infty}xe^{-\frac{x^2}{2}}\cdot\ \sum_{k=0}^{\infty} \frac{x^{2k}}{2^{2k}{k!}^2}\mathrm{d}x = \sum_{k=0}^{\infty} \left( \frac{1}{2^{2k}{k!}^2}\int_{0}^{\infty}e^{-\frac{x^2}{2}}\cdot\  x^{2k+1}\mathrm{d}x \right)
$$
Know with the substitution $x\to \sqrt{2x}$ we obtain:
$$I=\sum_{k=0}^{\infty} \left( \frac{1}{2^{2k}{k!}^2}\int_{0}^{\infty}\frac{\sqrt2}{2\sqrt{x}}\cdot e^{-x}\cdot\  2^{k+\frac{1}{2}}\cdot x^{k+\frac{1}{2}}\mathrm{d}x \right)=\sum_{k=0}^{\infty} \left( \frac{1}{2^{k}{k!}^2}\int_{0}^{\infty} e^{-x}\cdot\ x^{k}\mathrm{d}x \right)=\sum_{k=0}^{\infty} \frac{1}{2^{k}{k!}}=\sqrt e$$
Where I used that by the integral definition of the gamma function $\int_{0}^{\infty} e^{-x}\cdot\ x^{k}\mathrm{d}x=\Gamma (k+1)=k!$ and the well known series expansion $e^x=\sum_{k=0}^{\infty} \frac{x^k}{{k!}}$.
