Understand integral from Gradshteyn and Ryzhik book "Table of integrals, series, products" I was checking useful integrals in this book. I have found one (6.298) that is what I need, but I don't understand how every step towards the final result works. 
$$\int_0^{+\infty}\,\left[2\cosh(ab)-e^{-ab}\Phi\left(\frac{b-2ax^2}{2x}\right)-e^{ab}\Phi\left(\frac{b+2ax^2}{2x}\right)\right]\,x\,e^{-(\mu-a^2)x^2}\,\,dx=\frac{1}{\mu-a^2}e^{-b\sqrt{\mu}}$$
where $\Phi(x)=erf(x)$, $a,b>0$ and $Re\,\mu>0$.
Can anybody help me with the intermediate steps to get the final result?
Are there other conditions missing? Like $\mu-a^2>0$?
 A: This example is easier than I thought. Define:
\begin{equation}
f^{(a,b)}(x):= 2 \cosh(a b)- \exp(-a b) Erf[\frac{b- 2 a x^2}{2 x}] - \exp(a b) Erf[\frac{b+ 2 a x^2}{2 x}]
\end{equation}
Then 
\begin{eqnarray}
&&\int\limits_0^\infty f^{(a,b)}(x) \cdot x \exp\left(-(\mu-a^2) x^2\right) dx=\\
&&
\int\limits_0^\infty \frac{1}{2} \frac{2 b e^{-\frac{4 a^2 x^4+b^2}{4 x^2}}}{\sqrt{\pi } x^2} \cdot \frac{\exp\left(-(\mu-a^2) x^2\right)}{\mu-a^2} dx=\\
&&\frac{b}{\sqrt{\pi }\left(\mu -a^2\right)} \int\limits_0^\infty\frac{e^{-\frac{b^2+4 \mu  x^4}{4 x^2}}}{ x^2 } dx \quad (i)
\end{eqnarray}
where in the second line we integrated by parts and in the third line we simplified the result.
Now we have:
\begin{eqnarray}
(-b) \frac{e^{-\frac{b^2+4 \mu  x^4}{4 x^2}}}{ x^2 } = 
\left(-\frac{b}{2 x^2}+\sqrt{\mu}\right) e^{-\left(\frac{b}{2 x} + \sqrt{\mu} x\right)^2+b\sqrt{\mu}} +
\left(-\frac{b}{2 x^2}-\sqrt{\mu}\right) e^{-\left(\frac{b}{2 x} - \sqrt{\mu} x\right)^2-b\sqrt{\mu}}
\end{eqnarray}
Therefore by integrating both sides we get:
\begin{eqnarray}
(-b) \int \frac{e^{-\frac{b^2+4 \mu  x^4}{4 x^2}}}{ x^2 } dx = 
e^{b \sqrt{\mu}} \frac{\sqrt{\pi}}{2} Erf[\frac{b}{2 x}+\sqrt{\mu} x] + e^{-b\sqrt{\mu}} \frac{\sqrt{\pi}}{2} Erf[\frac{b}{2 x}-\sqrt{\mu} x]
\end{eqnarray}
Therefore the definite integral reads:
\begin{eqnarray}
(-b) \int\limits_0^\infty \frac{e^{-\frac{b^2+4 \mu  x^4}{4 x^2}}}{ x^2 } dx = 
e^{-b\sqrt{\mu}} \frac{\sqrt{\pi}}{2} (-2) \quad (ii)
\end{eqnarray}
because there is only a contribution from the second error function on the right hand side. Now $(i)$ along with $(ii)$ yields the required result.
