How big is the integral $\int_0^\infty \frac{x\exp(-x^2/4)\cosh(x)}{\sqrt{\cosh(x)-1}} dx$ I can't seem to get Maple to approximate the integral 
$$\int_0^\infty \frac{x\exp(-x^2/4)\cosh(x)}{\sqrt{\cosh(x)-1}} dx.$$
Could somebody tell me why?
This integral "should be" well-defined. (My reasons are not mathematical. The  book I'm reading suggest that this integral makes sense.) Do note that the denominator of the integrand explodes at $x=0$, but this should not be a problem... 
Can we give an upper bound for this integral?
 A: Using the following inequality:
$$
  \frac{x}{\sqrt{\cosh{x}-1}} = \sqrt{2} \frac{x/2}{\sinh(x/2)} \leqslant \sqrt{2}
$$
It is easy to work out the upper bound:
$$
 \int_0^\infty \frac{x\exp(-x^2/4)\cosh(x)}{\sqrt{\cosh(x)-1}} \mathrm{d}x <
   \sqrt{2} \int_0^\infty \exp(-x^2/4)\cosh(x) \mathrm{d}x = \sqrt{2 \pi} \mathrm{e} \approx 6.8 
$$
A: With a little bit more work you can use the inequality given by Sasha to get a tighter upper bound. Indeed,
$$
\frac{\cosh x}{\sqrt{\cosh x -1}}=\frac{1+2\,\sinh^2(x/2)}{\sqrt{2}\sinh(x/2)}=\frac{1}{\sqrt{2}\sinh(x/2)}+\sqrt{2}\sinh(x/2),
$$
which splits the original integral into two terms. The first of them may be estimated using Sasha's inequality:
$$
\int_0^{\infty} \frac{x\exp(-x^2/4)}{\sqrt{2}\sinh(x/2)}dx=\sqrt{2}\int_0^{\infty} \frac{(x/2)\exp(-x^2/4)}{\sinh(x/2)}dx<\sqrt{2}\int_0^{\infty} \exp(-x^2/4)dx=\sqrt{2\pi}.
$$
The second integral computes exactly:
$$
\sqrt{2}\int_0^{\infty} x\exp(-x^2/4)\sinh(x/2) dx=\sqrt{2\pi}\mbox{e}^{1/4},
$$
(see section 3.562 in Gradshteyn & Ryzhik's "Table of Integrals, Series and Products" or simply use Mathematica or Maple). Therefore, we obtain
$$
\int_0^{\infty} \frac{x\exp(-x^2/4)\cosh x}{\sqrt{\cosh x -1}}dx<\sqrt{2\pi}(1+\mbox{e}^{1/4})\approx 5.7252.
$$
