Evaluation of tricky integral I want to evaluate the integral
$$\int _ {b} ^ {\infty} \mathrm{d} x \, \frac{e ^ {x ^ {2} / s} (b^2 + 3 x ^ 2) ^ {2}}{x (x^2 + b^2)}$$,
where $b$ and $s$ are positive real numbers. I thought of writing $x$ as $x = - i y$. Then $y$ would be a pure imaginary number and the integral would become
$$\int _ {i b} ^ {i \infty} \mathrm{d} y \, \frac{e ^ {- y ^ {2} / s} (b^2 - 3 y ^ 2) ^ {2}}{y (- y^2 + b^2)}$$.
I tried then to do the integral by contour integration and consider the integral
$$\int  \mathrm{d} z \, \frac{e ^ {- z ^ {2} / s} (b^2 - 3 z ^ 2) ^ {2} \log(z - i b)}{z (- z^2 + b^2)}$$,
with a branch cut from $ib$ to $i \infty$. Am I on the right track? Could someone please help me?
 A: As proposed the integral has a issue with large values, ie $x \rightarrow \infty$, since the integral
is unbounded. This can be remedied by replacing $s$ with $- 1/s$. In this view consider the integral
\begin{align}
I = \int_{b}^{\infty} e^{-s x^{2}} \ \frac{(a x^{2} + b^{2})^{2}}{x ( x^{2} + b^{2}) } \ dx.
\end{align}
It is readily seen that
\begin{align}
\frac{(a x^{2} + b^{2})^{2}}{x ( x^{2} + b^{2}) } &= \frac{ [a (x^{2}+b^{2}) + (1-a) b^{2}]^{2}}
{x(x^{2}+b^{2})} \\
&= \frac{a^{2} (x^{2}+b^{2})^{2} + 2 a (1-a) b^{2} (x^{2}+b^{2}) + (1-a)^{2}b^{4}}{x(x^{2}+b^{2})} \\
&= a^{2} \ \frac{x^{2}+b^{2}}{x} + 2 a(1-a) b^{2} \ \frac{1}{x} + \frac{(1-a)^{2} b^{4}}{x(x^{2}
+ b^{2})} \\
&= a^{2} x + \frac{a(2-a)b^{2}}{x} + \frac{(1-a)^{2} b^{4}}{x(x^{2}+ b^{2})}
\end{align}
for which the integral becomes
\begin{align}
I = a^{2} \int_{b}^{\infty} e^{-s x^{2}} x \ dx + a(2-a)b^{2} \int_{b}^{\infty} e^{-s x^{2}} \ 
\frac{dx}{x} + (1-a)^{2} \int_{b}^{\infty} \frac{e^{-s x^{2}} \ dx}{x(x^{2} + b^{2})}.
\end{align}
Let $x = \sqrt{t/s}$ in each of the integrals to obtain
\begin{align}
I = \frac{a^{2}}{2s} \ e^{- s b^{2}} + \frac{a(2-a)b^{2}}{2} \int_{sb^{2}}^{\infty} \frac{e^{-t} \ dt}
{t} + \frac{s (1-a)^{2}}{2} \int_{sb^{2}}^{\infty} \frac{e^{-t} \ dt}{t(t+sb^{2})}.
\end{align}
The last integral can be reduced in the following. Using
\begin{align}
\frac{1}{t(t+sb^{2})} = \frac{1}{s b^{2}} \left( \frac{1}{t} - \frac{1}{t+s b^{2}} \right)
\end{align}
then
\begin{align}
\int_{sb^{2}}^{\infty} \frac{e^{-t} \ dt}{t(t+sb^{2})} &= \frac{1}{s b^{2}} \int_{sb^{2}}^{\infty}
e^{-t} \left( \frac{1}{t} - \frac{1}{t+s b^{2}} \right) dt \\
&= \frac{1}{s b^{2}} \int_{sb^{2}}^{\infty} \frac{e^{-t} \ dt}{t} - \frac{1}{s b^{2}} \int_{sb^{2}
}^{\infty} \frac{e^{-t} \ dt}{t+s b^{2}} \\
&= \frac{1}{s b^{2}} \int_{sb^{2}}^{\infty} \frac{e^{-t} \ dt}{t} - \frac{e^{s b^{2}}}{s b^{2}}
\int_{2 s b^{2}}^{\infty} \frac{ e^{-t} \ dt}{t}.
\end{align}
Using this reduction in the integral $I$ leads to
\begin{align}
I = \frac{a^{2}}{2s} \ e^{- s b^{2}} + \frac{a(2-a)b^{4}+(1-a)^{2}}{2 b^{2}} \int_{sb^{2}}^{\infty}
\frac{e^{-t} \ dt}{t} - \frac{(1-a)^{2}}{2 b^{2}} \ e^{s b^{2}} \int_{2 s b^{2}}^{\infty} 
\frac{e^{-t} \ dt}{t}.
\end{align}
The exponential integral $E_{1}(z)$ is given by
\begin{align}
E_{1}(z) = \int_{z}^{\infty} \frac{e^{-t}}{t} \ dt
\end{align}
and can be used to reduce the integral in question to the value
\begin{align}
I = \frac{a^{2}}{2s} \ e^{- s b^{2}} + \frac{a(2-a)b^{4}+(1-a)^{2}}{2 b^{2}} E_{1}(s b^{2})
- \frac{(1-a)^{2}}{2 b^{2}} \ e^{s b^{2}} E_{1}(2 s b^{2}).
\end{align}
Hence
\begin{align*}
& \int_{b}^{\infty} e^{-s x^{2}} \ \frac{(a x^{2} + b^{2})^{2}}{x ( x^{2} + b^{2}) } \ dx \\
& \hspace{5mm} = \frac{a^{2}}{2s} \ e^{- s b^{2}} + \frac{a(2-a)b^{4}+(1-a)^{2}}{2 b^{2}} E_{1}(s b^{2})
- \frac{(1-a)^{2}}{2 b^{2}} \ e^{s b^{2}} E_{1}(2 s b^{2}).
\end{align*}
