I've been trying to solve the definite integral
\begin{align*}
I = \int_{0}^{u} e^{-\sqrt{a^2 – b^2} \cosh(x)} \, dx \, ,
\end{align*}
with $u = \mathrm{arctanh}(\frac{b}{a})$ and $a > b, \, a > 0$. Performing the change of variables $x = \mathrm{arccosh}(y)$, we can rewrite the integral
\begin{align*}
I = \int_{1}^{\cosh(u)} \frac{1}{\sqrt{y^2 – 1}} e^{-\sqrt{a^2 – b^2} \, y} \, dy \, .
\end{align*}
I feel like it should be possible to evaluate this, and my intuition tells me that the indefinite integral will come out as an error function since it looks very similar to
\begin{align*}
\int \frac{1}{\sqrt{x}} e^{-\alpha x} \, dx = \sqrt{\frac{\pi}{\alpha}} \mathrm{erf}(\sqrt{\alpha x})
\end{align*}
for $\alpha > 0$. I've looked through Gradshteyn and Ryzhik, and Prudnikov, Brychkov, and Marichev, but haven't found anything quite what I'm looking for. Does anyone else know how to approach this integral?
P.S. The original integral was
\begin{align*}
\int_0^\infty dx \, e^{-\sqrt{a^2 – b^2} \cosh(x – u)} \, ,
\end{align*}
which I split up into something that gives a modified Bessel function of the second kind, and the other term is the integral I have above.
Best Answer
So it looks like Gary's right, and the same integral was also identified as an incomplete modified Bessel function in a different answer here by Paul. Some relevant papers are: Jones, Cicchetti and Faraone, and Shu and Shastri.