Value of $\int _0^{+\infty }\:e^{-x²}\cdot \sqrt{\sinh^2\left(x\right)+1}dx$ Can somebody help me find the value of
$\int _0^{+\infty }\:e^{-x^2}\cdot \sqrt{\sinh^2\left(x\right)+1}dx$
without using any special integral (so no erf)
I keep getting stuck trying to solve the integral with the following integral
$\int _0^{+\infty }\:\frac{\sqrt{e^{-2x}+e^{2x}+2}}{2e^{x^2}}dx$
a different method gave me
$\int _0^{+\infty }\:\left(e^{-x^2+x}+e^{-x^2-x}\right)dx$
 A: $\textbf{Hint}$: The function is even therefore
$$I = \frac{1}{2}\int_{-\infty}^\infty e^{-x^2}\cosh x\:dx = \frac{1}{2}\int_{-\infty}^\infty e^{-x^2+x}\:dx$$
Can you continue from here?
A: HINT
As suggested by @ThomasAndrews, one gets
\begin{align*}
\exp(-x^{2})\sqrt{\sinh^{2}(x) + 1} & = \exp(-x^{2})\cosh(x)\\\\
& = \exp(-x^{2})\left(\frac{\exp(x) + \exp(-x)}{2}\right)\\\\
& = \frac{\exp(-x^{2} + x) + \exp(-x^{2} - x)}{2}\\\\
& = \frac{1}{2}\exp\left(\frac{1}{4}\right)\left[\exp\left(-\left(x - \frac{1}{2}\right)^{2}\right) + \exp\left(-\left(x + \frac{1}{2}\right)^{2}\right)\right]
\end{align*}
Can you proceed from here?
A: $\newcommand{\d}{\,\mathrm{d}}$Since the integrand is even and the improper integrals clearly converge: $$\int_0^\infty e^{-x^2}\cosh(x)\d x=\frac{1}{2}\int_{-\infty}^\infty e^{-x^2}\cosh(x)\d x$$And: $$\begin{align}\int_{-\infty}^\infty e^{-x^2}\cosh(x)\d x&=\frac{1}{2}\left[\int_{-\infty}^\infty e^{x-x^2}\d x+\int_{-\infty}^\infty e^{-(x^2+x)}\d x\right]\\&=\frac{e^{1/4}}{2}\left[\int_{-\infty}^\infty e^{-(x-1/2)^2}\d x+\int_{-\infty}^\infty e^{-(x+1/2)^2}\d x\right]\\&=\frac{e^{1/4}}{2}\left[\int_{-\infty}^\infty e^{-x^2}\d x+\int_{-\infty}^\infty e^{-x^2}\d x\right]\\&=e^{1/4}\sqrt{\pi}\end{align}$$
So you have: $$\int_0^\infty e^{-x^2}\cosh(x)\d x=\frac{1}{2}e^{1/4}\sqrt{\pi}$$
