Integral $\int_0^\infty \frac{xe^{-bx}}{\sqrt{x^2 +a^2}}dx$

Question

Is the following integral solvable analytically? $a$ and $b$ are constants. $$ \int_0^\infty \frac{xe^{-bx}}{\sqrt{x^2 +a^2}}dx $$

Answer 1

Letting $x=a\sinh t$ or $x=\dfrac a2~\sinh t$, and employing the integral expressions for the Bessel and Struve functions, we ultimately arrive at $\displaystyle\int_0^\infty\dfrac u{\sqrt{u^2+1}}~e^{-cu}~du~=~\int_0^\infty\sinh t\cdot e^{-c\sinh t}~dt~=$ $=-\dfrac\pi2\Big[Y_1(c)+H_{-1}(c)\Big]$, where c is a parameter whose expression depends on a and b.

Answer 2

This will be a very modest contribution after Lucian's answer.

Starting with $$I=\int_0^\infty \frac{xe^{-bx}}{\sqrt{x^2 +a^2}}dx$$ and changing variable $x=a y$, we have $$I=a\int_0^\infty \frac{y\, e^{-c y}}{\sqrt{y^2+1}}dy$$ with $c=ab$. We can perform a Taylor expansion and get $$\frac{y}{\sqrt{y^2+1}}=y-\frac{y^3}{2}+\frac{3 y^5}{8}-\frac{5 y^7}{16}+\frac{35 y^9}{128}-\frac{63 y^{11}}{256}+O\left(y^{13}\right)=\sum_{i=1}^{\infty}a_iy^i$$ and we are just left with integrals $$J_n=\int_0^\infty y^n e^{-cy}dy=\frac{n!}{c^{n+1}}$$

Edit : Robjohn's comment "This will only give an asymptotic series in $c$ since the radius of convergence is $0$" must be very seriously taken into account.

Answer 3

(answer requires analysis of one complex variable)

Let us first replace the numerator with $|x|e^{-b|x|}$ as this would not make any difference for our problem. Now create a half circle, positive imaginary half plane, centered at origo and let radius grow without bounds. Obviously at the arc, our function value would diminish much faster than the radius grows. Then let us observe that our function is even with respect to the real part of x. So the value of our contour integral is $2I$ where $I$ is the integral we seek. Furthermore we see that there are two poles of our function, located at $x = \pm ai$. However only the positive one is inside our contour.

Now we can use the residue theorem to calculate 2I. That can be left as an exercise :)