Are there any nice expressions for $\int_0^\infty e^{-x^2}\sqrt{x^2-k^2}\ \mathrm{d}x$?

Question

In some applied mathematics (ocean modelling) I was doing I came across the integral

$$I(k)=\int_0^\infty e^{-x^2}\sqrt{x^2-k^2}\ \mathrm{d}x,$$

where $k\geq 0$ is a constant that depends on the parameters (e.g. flow speed) I wish to use. I was wondering whether anyone knows of a neat expression for this integral, even in terms of Bessel functions? I'm particularly interested in the real part of this integral, that is

$$\Re(I(k))=\int_k^\infty e^{-x^2}\sqrt{x^2-k^2}\ \mathrm{d}x.$$

I was hoping for a solution similar this question but have had no luck. An asymptotic expression (for large $k$) for $I(k)$ or $\Re(I(k))$ would also be really useful.

Answer 1

Substituting $$x = k \cosh t, \qquad dx = k \sinh t \,dt$$ transforms the integral for $\Re(I(k))$ to $$k^2 \int_0^\infty \sinh^2 t e^{-k^2 \cosh^2 t} \,dt = \frac{1}{4} k^2 \exp \left(-\frac{k^2}{2}\right) \left[K_1\left(\frac{k^2}{2}\right) - K_0\left(\frac{k^2}{2}\right)\right],$$ where $K_\alpha$ is the modified Bessel function of the second kind. Here the equality follows from hyperbolic trigonometric identities and the integral representation $$K_\alpha(s) = \int_0^\infty e^{-s \cosh t} \cosh \alpha t \,dt .$$ Using known expansions for Bessel functions gives that:

• as $k \to \infty$, $$\Re(I(k)) = \frac{\sqrt{\pi}} {4} \cdot \frac{1}{k} e^{-k^2} \left(1 - \frac{3}{4 k^2} + R_1(k) \right),$$ where $R_1(k) \in O\left(\frac{1}{k^4}\right)$, and
• as $k \to 0$, $$\Re(I(k)) = \frac{1}{2} \left[1 + k^2 \log |k| + \left(-\frac{1}{2} + \log 2 + \frac{1}{2} \gamma\right) k^2 + R_2(k)\right],$$ where $R_2(k) \in O(k^4 \log |k|)$ and $\gamma$ is the Euler-Mascheroni constant.

Similarly, the imaginary part of the integral is $$\Im (I(k)) = \int_0^k e^{-x^2} \sqrt{k^2 - x^2} \,dx = \frac{1}{4} \pi k^2 \exp \left(-\frac{k^2}{2}\right) \left[I_0\left(\frac{k^2}{2}\right) + I_1\left(\frac{k^2}{2}\right)\right] ,$$ where $I_0$ is the modified Bessel function of the first kind. (Here we've used the principal branch of the square root function; a different choice of branch may yield the negative of this quantity.) Again using known asymptotics for $I_0$ gives asymptotic expansions for $\Im(I(k))$:

• As $k \to \infty$, $$\Im(I(k)) = \frac{\sqrt{\pi}}{2} k \left(1 - \frac{1}{4 k^2} + R_3(k)\right) ,$$ where $R_3(k) \in O\left(\frac{1}{k^4}\right)$, and
• as $k \to 0$, $$\Im(I(k)) = \frac{\pi}{4} k^2 \left(1 - \frac{1}{4} k^2 + R_4(k)\right) ,$$ where $R_4(k) \in O(k^4)$, and

(The leading term of the small-$k$ expansion for $\Im(I(k))$ needs nothing sophisticated, by the way, just the evenness of the integral in $k$ and the formula for the area of a circle.)

Answer 2

Consider the limit

$$\lim_{k\to\infty}ke^{k^2}\int_k^\infty dx\:e^{-x^2}\sqrt{x^2-k^2}$$

and use the substitution $t^2 = x^2-k^2$

$$\lim_{k\to\infty}\int_0^\infty dt\:e^{-t^2}\frac{kt^2}{\sqrt{t^2+k^2}}\longrightarrow \int_0^\infty dt\:t^2e^{-t^2} = \frac{\sqrt{\pi}}{4}$$

by dominated convergence. Therefore

$$\operatorname{Re}\{I(k)\} \sim e^{-k^2}\left[\frac{\sqrt{\pi}}{4k} + O\left(\frac{1}{k^2}\right)\right]$$

Answer 3

Using Meijer G-function $$I(k)=\int_{\color{red}{\large k}}^\infty e^{-x^2}\,\sqrt{x^2-k^2}\,dx=\frac{\sqrt{\pi }}{4}\, k\,G_{1,2}^{2,0}\left(k^2| \begin{array}{c} 1 \\ -\frac{1}{2},\frac{1}{2} \\ \end{array} \right)$$

Expanding for large values of $k$ $$I(k)=\frac{\sqrt{\pi }}{4 k}\, e^{-k^2}\left(1-\frac{3}{4 k^2}+\frac{45}{32 k^4}-\frac{525}{128 k^6}+O\left(\frac{1}{k^8}\right) \right)$$ which more than decent for $k>3$.

Using @Travis Willse's result for the missing part $$\Re\Bigg( \int_{0}^\infty e^{-x^2}\,\sqrt{x^2-k^2}\,dx\Bigg)=$$ $$\frac{\sqrt \pi}{4k} \,e^{-k^2}\,\left(1-\frac{3}{4 k^2}+\frac{45}{32 k^4}-\frac{525}{128 k^6}+O\left(\frac{1}{k^8}\right) \right)$$ which seems to be quite decent even for small values of $k$.

In a logarithmic scale

$$\left( \begin{array}{ccc} k & \text{approximation} & \text{exact} \\ 2 & -5.6858400 & -5.6507359 \\ 3 & -10.986833 & -10.985083 \\ 4 & -18.243531 & -18.243333 \\ 5 & -27.451780 & -27.451744 \\ 6 & -38.625724 & -38.625716 \\ 7 & -51.774705 & -51.774702 \\ 8 & -66.904827 & -66.904827 \\ \end{array} \right)$$

Edit

If you want a better approximation, transforming the series expansion into a simple $[2n,2n]$ Padé approximant $$\frac{\sqrt \pi}{4k} \,e^{-k^2}\,\frac{320 k^4+1608 k^2+779}{320 k^4+1848 k^2+1715}$$