TITLE: Integral $\int_0^\infty\left(x+5\,x^5\right)\operatorname{erfc}\left(x+x^5\right)\,dx$
QUESTION [41 upvotes]: Is it possible to find a closed form (possibly using known special functions) for this integral?
$$\int_0^\infty\left(5\,x^5+x\right)\operatorname{erfc}\left(x^5+x\right)\,dx$$
where $\operatorname{erfc}$ is the complementary error function
$$\operatorname{erfc} x=\frac{2}{\sqrt{\pi}}\int_x^{\infty}e^{-z^2}dz.$$

REPLY [10 votes]: Hint:
$\int_0^\infty(5x^5+x)\text{erfc}(x^5+x)~dx$
$=\int_0^\infty\text{erfc}(x^5+x)~d\left(\dfrac{5x^6}{6}+\dfrac{x^2}{2}\right)$
$=\left[\left(\dfrac{5x^6}{6}+\dfrac{x^2}{2}\right)\text{erfc}(x^5+x)\right]_0^\infty-\int_0^\infty\left(\dfrac{5x^6}{6}+\dfrac{x^2}{2}\right)d(\text{erfc}(x^5+x))$
$=\int_0^\infty\left(\dfrac{5x^6}{6}+\dfrac{x^2}{2}\right)(5x^4+1)\dfrac{2e^{-x^2(x^8+1)}}{\sqrt\pi}dx$
$=\dfrac{2}{\sqrt\pi}\int_0^\infty e^{-(\sinh^\frac{1}{2}x)(\sinh^2x+1)}\left(\dfrac{5\sinh^\frac{3}{2}x}{6}+\dfrac{\sinh^\frac{1}{2}x}{2}\right)(5\sinh x+1)~d(\sinh^\frac{1}{4}x)$
$=\dfrac{1}{2\sqrt\pi}\int_0^\infty e^{-\sinh^\frac{1}{2}x\cosh^2x}~\left(\dfrac{5\sinh^\frac{3}{2}x}{6}+\dfrac{\sinh^\frac{1}{2}x}{2}\right)(5\sinh x+1)\sinh^{-\frac{3}{4}}x\cosh x~dx$
That's why it may relate to Bessel function.