Given $c>0$, can the integral $$ \int_{0}^{1} \frac{x e^{-cx}}{\sqrt{1-x^2}}\,dx $$ be expressed in closed form where special functions are allowed? (I can evaluate it numerically for any given value of $c$. But if possible I'm seeking an expression around $c$, and preferably an expression that I could use in a single cell in a spreadsheet.)
My attempted solution: Let $x = \cos\theta$. Then $$\begin{align*} \int_{0}^{1} \frac{x e^{-cx}}{\sqrt{1-x^2}}\,dx &= \int_{\frac{\pi}{2}}^{0} \frac{e^{-c\cos\theta} \cos\theta}{\sin\theta} (-\sin\theta) \, d\theta \\&= \int_{0}^{\frac{\pi}{2}} e^{-c\cos\theta} \cos\theta \, d\theta \end{align*}$$ Now for any integer $n$ we have $$ I_n(z) = \int_{0}^{\pi} e^{z\cos\theta} \cos(n\theta) \, d\theta $$ where $I_n(z)$ is the modified Bessel function of the first kind. Putting $n=1$ matches the integrands but not the limits of integration.
Many thanks in advance for any suggestions.
There's a combination of a modified Bessel function and something else since the structure of the integral $\int_{0}^{\pi/2}\left(\cos\theta\right)^n\,d\theta$ depends on the parity of $n$. Namely we have
$$ \int_{0}^{\pi/2}\left(\cos\theta\right)^{2n}\,d\theta = \frac{\pi}{2\cdot 4^n}\binom{2n}{n},\qquad \int_{0}^{\pi/2}\left(\cos\theta\right)^{2n+1}\,d\theta = \frac{4^n}{(2n+1)\binom{2n}{n}}$$ so
$$\begin{eqnarray*} \int_{0}^{1}\frac{e^{-cx}}{\sqrt{1-x^2}}\,dx&=&\int_{0}^{\pi/2}e^{-c\cos\theta}\,d\theta=\sum_{n\geq 0}\frac{(-1)^n c^n}{n!}\int_{0}^{\pi/2}\left(\cos\theta\right)^n\,d\theta\\&=&\frac{\pi}{2}\sum_{n\geq 0}\frac{c^{2n}\binom{2n}{n}}{(2n)!4^n}-\sum_{n\geq 0}\frac{c^{2n+1}4^n}{(2n+1)!(2n+1)\binom{2n}{n}}\\[0.2cm]&=&\frac{\pi}{2}\underbrace{\sum_{n\geq 0}\frac{c^{2n}}{4^n n!^2}}_{\text{Bessel}}-\underbrace{\sum_{n\geq 0}\frac{c^{2n+1}4^n n!^2}{(2n+1)!^2}}_{\text{Not Bessel}}\end{eqnarray*} $$ and the same method applies to $ \int_{0}^{1}\frac{x e^{-cx}}{\sqrt{1-x^2}}\,dx $, which is just the opposite of the derivative of the last series:
$$ \int_{0}^{1}\frac{x e^{-cx}}{\sqrt{1-x^2}}\,dx = \underbrace{\sum_{n\geq 0}\frac{c^{2n}4^n n!^2}{(2n)!(2n+1)!}}_{\text{Not Bessel}}-\frac{\pi}{4}\underbrace{\sum_{n\geq 0}\frac{c^{2n+1}}{4^n n!(n+1)!}}_{\text{Bessel}}.$$