Integral: $\int_{1}^{\infty} \operatorname{arctanh} \left(k\,\sqrt{\frac{x^2-1}{x^2}} \right)\, e^{-\alpha \, x \,} dx$ Is there any chance to find a closed form for these integrals?
$$I_1(k,\alpha)=\int_{1}^{\infty} \operatorname{arctanh}  \left(k\,\sqrt{\frac{x^2-1}{x^2}} \right)\, e^{-\alpha \, x \,} dx$$
$$I_2(k,\alpha)=\int_{1}^{\infty} x^2\, \operatorname{arctanh}  \left(k\,\sqrt{\frac{x^2-1}{x^2}} \right)\, e^{-\alpha \, x} \, dx$$
$\operatorname{arctanh} (z)=\tanh^{-1}(z)$ is the inverse hyperbolic tangent.
$k$ and $\alpha$ are real constants.
These two integrals appear in part of my master's work in plasma physics. Obviously, solutions are calculated numerically. But the symbolic results allow me to discuss the solutions of an algebraic equation. Surely there's a chance that someone might be able to calculate them.
 A: After the integration by parts and change of variable
$$1-\frac{1}{x^2}=t^2$$
we have
$$I_1(k,\alpha)=\frac{k}{\alpha}\int_{0}^{1}\frac{e^{-\frac{\alpha}{\sqrt{1-t^2}}}}{1-k^2t^2}dt$$
We note that in order for the integral to make sense, inequality
$$\left| k \right|\leqslant 1$$
must apply.
This integral probably does not have a manageable closed form expression.
In fact, to proceed further, physical considerations should be included.
It remains only to try to find an approximate analytical expression based on mathematical considerations only.
Let's analyze the integrand.
If $\alpha$ is large enough, the integrand decreases quickly and its max value is reached at $t=0$.
In that case we have
$$I_1(k,\alpha)<k\frac{e^{-\alpha}}{\alpha}$$
If $\alpha$ is small enough then the integrand has a maximum within the interval $\left  [ 0 , 1\right ]$ and the equality above may not apply.
But despite that, we consider it to be a reasonable approximation.
$$I_1(k,\alpha)\approx I_{1approx}(k,\alpha)=k\frac{e^{-\alpha}}{\alpha}$$
To decide is the approximation good, bad or acceptable, below is few examples.
$k=1,\alpha=1$
$$I_1(k,\alpha)=0.421...$$
$$I_{1approx}(k,\alpha)=0.367...$$
$k=0.77,\alpha=2$
$$I_1(k,\alpha)=0.0374...$$
$$I_{1approx}(k,\alpha)=0.0521...$$
$k=0.77,\alpha=0.5$
$$I_1(k,\alpha)=0.960...$$
$$I_{1approx}(k,\alpha)=0.934...$$
Updates after receiving new information
The new information is that $\alpha$ varies from 0.001 to 1000.
For very large $\alpha$ the approximation above still applies.
Example
$k=1,\alpha=10$
$$I_1(k,\alpha)=0.000002$$
$$I_{1approx}(k,\alpha)=0.000005$$
For very small $\alpha$, i.e. $\alpha \ll 1$ we expand the exponent in the integrand into Taylor series
$$e^{-\frac{\alpha}{\sqrt{1-t^2}}}=1-\frac{\alpha}{\sqrt{1-t^2}}+...$$
Keeping only the first two terms we get after elementary integration the following approximation
$$I_{1approx}(k,\alpha)=\frac{1}{2\alpha}\ln \frac{1+k}{1-k}-\frac{\pi}{2}\frac{k}{\sqrt{1-k^2}}$$
Examples
$k=0.9,\alpha=0.01$
$$I_1(k,\alpha)=144.095$$
$$I_{1approx}(k,\alpha)=143.978$$
$k=0.9,\alpha=0.001$
$$I_1(k,\alpha)=1468.99$$
$$I_{1approx}(k,\alpha)=1468.98$$
It is important to note that the case $k=1$  does not apply here. A separate study is required for this.
A: Following  the integral Martin Gales provided, it is possible to evaluate by the Laplace method for large values of $\alpha$. I.e. I'm only looking at this specific case, not an exact solution or the behavior of small $\alpha$.
Following the procedure in the link, the integral may then be written as
\begin{equation}
\int_0^1 (1-k^2t^2)^{-1}\exp\left( -\alpha(1-t^2)^{-1/2}\right) dt = \int_0^1 q(t) \exp(-\alpha p(t)) dt
\end{equation}
The minimum is obtained or $t=0$ for which one may calculate that
\begin{align}
p(t) = & (1-t^2)^{-1/2} = \sum_{n=0}^{\infty} {n-1/2\choose n}t^{2n} = 1+\frac{1}{2}t^2 + ... \newline
q(t) = & (1-k^2t^2)^{-1} = \sum_{n=0}^{\infty}(kt)^{2n} = 1 + k^2t^2 + ...
\end{align}
for which one may identify $p(0)=1,\mu =2, p_0 = 1/2 $ and $\lambda = 1, q_0 = 1$. Then, to leading order one has that
\begin{equation}
\int_0^1 (1-k^2t^2)^{-1}\exp\left( -a(1-t^2)^{-1/2}\right) dt \sim e^{-\alpha p(0)} \Gamma\left(\frac{\lambda}{\mu}\right) \frac{q_0}{\mu(p_0\alpha)^{\lambda/\mu}} = \sqrt{\frac{\pi}{2\alpha}}e^{-\alpha}
\end{equation}
which hence gives that $I_1(k,a) \sim \frac{k}{\alpha}\sqrt{\frac{\pi}{2\alpha}}e^{-\alpha}$. If necessary, one may calculate additional terms in the asymptotic expression. E.g., including the second term, one obtains that
\begin{equation}
I_1(k,\alpha) \sim \frac{k\sqrt{\pi}}{\alpha}e^{-\alpha} \left(\frac{1}{\sqrt{2\alpha}} - \frac{3}{64\alpha} \right) + O(\alpha^{-5/2}), \quad \alpha \rightarrow +\infty
\end{equation}
