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How would one go about evaluating the following integral?

$$ \int_{0}^{\infty} \frac{Q}{\sqrt{4\pi t^3}} \exp\Big(-\frac{(Q - \alpha t)^2}{4t} - t(U^2 - \alpha U) \Big) dt$$

We have that $Q, \alpha > 0$, while $U$ is complex. Mathematica actually evaluates the whole thing to simply $e^{Q(\alpha - U)}$, under the assumption that $\Re(2U-\alpha)^2 > 0$.

My first attempt was to define $f(Q)$ as the integral above and show that it satisfies the differential equation of the exponential function; after differentiating the integrand I don't really see a way to recombine the terms though. I get something akin to

$$f'(Q) = \frac{1}{Q} f(Q) + \frac{\alpha}{2} f(Q) - \int_{0}^{\infty} \frac{Q^2}{4\sqrt{\pi t^5}} \exp\Big(-\frac{(Q - \alpha t)^2}{4t} - t(U^2 - \alpha U) \Big) $$

which is nowhere near $(\alpha - U) f(Q)$.

I've also looked through various integral tables without success, I only found a reference that integrals of products of rational and exponential functions could be resolved as sums of the $\mathrm{Ei}$ function, but do not see how in this case.

Marc Vaisband
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    Integral representation of 2nd modified Bessel function $K_{\nu}$, order $\nu = 1/2$. Half-integer order implies represented in terms of exponential, rational function. Common for evanescent solution of Helmholtz equation, spherical coordinates. – K B Dave Mar 10 '20 at 14:19
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    You can calculate that integral by adapting this. – J.G. Mar 10 '20 at 14:34

1 Answers1

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Assume that $(2U - \alpha)^{2}$ is positive.

A more convenient way to express the integral is $$ I =\frac{Q}{2 \sqrt{\pi}} \, \exp \left(\frac{\alpha Q}{2} \right) \int_{0}^{\infty}\exp \left(- \frac{\left(2U- \alpha\right)^{2}}{4}t- \frac{Q^{2}}{4t} \right) \, \frac{\mathrm dt}{t^{3/2}}. $$

Making the substitution $w = \frac{\left(2U- \alpha\right)^{2}}{4}t$, we get $$I = \frac{Q}{2 \sqrt{\pi}} \, \exp \left(\frac{\alpha Q}{2} \right) \frac{4}{(2U- \alpha)^{2}} \left( \frac{(2U - \alpha)^{2}}{4} \right)^{3/2}\int_{0}^{\infty} \exp \left(-w - \left(\frac{Q(2U- \alpha)}{2} \right)^{2}\frac{1}{4w}\right) \, \mathrm d w. $$

Then relating $I$ to a common integral representation of the modified Bessel function of the second kind, we get

$$\begin{align}I &= \frac{Q}{2 \sqrt{\pi}} \, \exp \left(\frac{\alpha Q}{2} \right) \frac{2U - \alpha}{2} \, 2 \left(\frac{Q(2U- \alpha)}{4} \right)^{-1/2} K_{1/2} \left(\frac{Q(2U- \alpha)}{2} \right) \\ &= \frac{Q}{2 \sqrt{\pi}} \, \exp \left(\frac{\alpha Q}{2} \right)(2U - \alpha) \left(\frac{Q(2U- \alpha)}{4} \right)^{-1/2} \sqrt{\frac{\pi}{2}} \, e^{-Q(U - \alpha/2)} \left( \frac{Q(2U- \alpha)}{2} \right)^{-1/2} \\ & = e^{Q(\alpha - U)}, \end{align} $$

where I used the fact that $K_{1/2}(z) = \sqrt{\frac{\pi}{2}} \frac{e^{-z}}{\sqrt{z}}$.

To argue that the result holds for $\Re(2U -\alpha)^{2} >0$, we can use the fact that the Laplace transform is analytic in the region of absolute convergence, and then appeal to the identity theorem.

Random Variable
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  • Thanks! That's super helpful. The NIST handbook actually lists the identity whenever $|\mathrm{ph}(z)| < \pi/4$, not just $z > 0$, so this might actually work even without the identity theorem, right? – Marc Vaisband Mar 11 '20 at 10:11
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    @MarcVaisband To get the integral in the correct form, I made a substitution. If I had assumed from the beginning that $U $ could be complex, the substitution would have transformed the integral from an integral along the positive real line to an integral somewhere else on the complex plane. – Random Variable Mar 11 '20 at 11:30
  • Of course, I'm sorry I'm blind – Marc Vaisband Mar 11 '20 at 12:13