I would like to solve the following integral

$$\int_0^{\infty} \frac{1}{t^{\frac{3}{2}}} e^{-\frac{a}{t}} \, \mathrm{erf}(\sqrt{t})\, \mathrm{d}t$$

with Re$(a)>0$ and erf the error function. Is it possible to given an closed form solution for this integral? Thank you.

Edit: Maybe this helps $$\mathrm{L}(\mathrm{erf}(\sqrt{t}),s)=\frac{1}{s \, \sqrt{1+s}}$$ $$\mathrm{L}^{-1}(t^{-\frac{3}{2}} e^{-\frac{a}{t}})=\frac{1}{\sqrt{\pi \, a}}\mathrm{sin}(2 \sqrt{a \, s})$$

with L the Laplace transform.

Therefore it should be $$\int_0^{\infty} \frac{1}{t^{\frac{3}{2}}} e^{-\frac{a}{t}} \, \mathrm{erf}(\sqrt{t})\, \mathrm{d}t = \int_0^{\infty} \frac{1}{s \, \sqrt{1+s} \, \sqrt{\pi \, a}} \, \mathrm{sin}(2 \sqrt{a \, s}) \mathrm{d}s$$


Represent the erf as an integral and work a substitution. To wit, the integral is

$$\frac{2}{\sqrt{\pi}} \int_0^1 dv \, \int_0^{\infty} \frac{dt}{t} e^{-(a/t+v^2 t)} $$

To evaluate the inner integral, we sub $y=a/t+v^2 t$. Then the reader can show that

$$\int_0^{\infty} \frac{dt}{t} e^{-(a/t+v^2 t)} = 2 \int_{2 v \sqrt{a}}^{\infty} \frac{dy}{\sqrt{y^2-4 a v^2}} e^{-y}$$

The latter integral is easily evaluated using the sub $y=2 v \sqrt{a} \cosh{w} $ and is equal to

$$2 \int_{2 v \sqrt{a}}^{\infty} \frac{dy}{\sqrt{y^2-4 a v^2}} e^{-y} = 2 \int_0^{\infty} dw \, e^{-2 v \sqrt{a} \cosh{w}} = 2 K_0 \left ( 2 v \sqrt{a} \right )$$

where $K_0$ is the modified Bessel function of the second kind of zeroth order. Now we integrate this expression with respect to $v$ and multiply by the factors outside the integral to get the final result:

$$\begin{align} \int_0^{\infty} dt \, t^{-3/2} e^{-a/t} \operatorname{erf}{\left ( \sqrt{t} \right )} &= \frac{4}{\sqrt{\pi}} \int_0^1 dv \, K_0 \left ( 2 v \sqrt{a} \right ) \\ &= 2 \sqrt{\pi} \left [K_0 \left ( 2 \sqrt{a} \right ) \mathbf{L}_{-1}\left ( 2 \sqrt{a} \right ) + K_1 \left ( 2 \sqrt{a} \right ) \mathbf{L}_{0}\left ( 2 \sqrt{a} \right ) \right ] \end{align}$$

where $\mathbf{L}$ is a Struve function.

  • $\begingroup$ Gorgeous. That modified Bessel Function completely surprised me... Bravo. $\endgroup$ – Brevan Ellefsen Aug 31 '16 at 12:48
  • $\begingroup$ Not that it's a big deal, but you could have instead made the substitution $t = \frac{\sqrt{a}}{v} e^{-u}$. $\endgroup$ – Random Variable Aug 31 '16 at 18:18
  • $\begingroup$ @RandomVariable: you're right - that would be a quicker way to the prize. Oh well, I still like my substitution from a graphical standpoint. $\endgroup$ – Ron Gordon Aug 31 '16 at 18:43
  • $\begingroup$ Could you probably explain the first substitution? I did not get how the denominator changes form $t^{3/2}$ to $t$. $\endgroup$ – Michael_K Sep 4 '16 at 14:59
  • $\begingroup$ @MK12 When you do the sun there is a factor of $\sqrt{t}$ outside the integral. $\endgroup$ – Ron Gordon Sep 4 '16 at 15:03

You can also follow your Laplace approach. Define $$ I(\alpha)=\int_0^{\infty}\frac{\sin(2\alpha\sqrt{s})}{s\sqrt{1+s}}ds $$

now set $s=q^2$

$$ I(\alpha)=2\int_0^{\infty}\frac{\sin(2\alpha q)}{q\sqrt{1+q^2}}ds $$

Now differentiate with respect to $\alpha$

$$ I'(\alpha)=4\int_0^{\infty}\frac{\cos(2\alpha q)}{\sqrt{1+q^2}}ds $$

this integral now furnishs a representation of the modified Besselfunction $K_0(z)$

$$ I'(\alpha)=4 K_0(2\alpha ) $$

according to 10.43.2 backintegrating w.r.t. to $\alpha$ yields

$$ I(\alpha)=2\pi \alpha (K_0(2\alpha )L_{-1}(2\alpha )-K_1(2\alpha )L_{0}(2\alpha ))+C $$

where $L_{\nu}(z)$ are modified Struve function. The constant of integration $C$ is fixed to be zero by the condition $I(0)=0$. Multiplying with $1/\sqrt{\pi}\alpha$ yields the result obtained by @Ron Gordon


Meh, interesting integral! I can give an heuristic approach but I believe someone else will do better. I'm on the bus and you know, it's not easy.

I would use Taylor Series for $e^{-a/t}$, hence

$$\int_0^{+\infty}\sum_{k = 0}^{+\infty} \frac{\left(-a/t\right)^k}{k!}t^{-3/2}\ \text{Erf}(\sqrt{t})\ \text{d}t$$

And we get

$$\sum_{k = 0}^{+\infty}\frac{(-a)^k}{k!}\int_0^{+\infty} t^{-k - 3/2}\ \text{Erf}(\sqrt{t})\ \text{d}t$$

Now if we call for simplicity $b = -k - 3/2$ we obtain a computable integral (I checked on Mathematica), which says:

$$\int_0^{+\infty} t^{b}\ \text{Erf}(\sqrt{t})\ \text{d}t = -\frac{\Gamma[3/2 + b]}{(1 + b)\sqrt{\pi}} ~~~ \to ~~~ -\frac{\Gamma[-k]}{(-k - 1/2)\sqrt{\pi}}$$

BUT there is condition over this result:

$$-\frac{3}{2} < \Re(b) < -1$$

This would give then

$$-\ \sum_{k = 0}^{+\infty}\frac{(-a)^k}{k!}\frac{\Gamma[-k]}{(-k - 1/2)\sqrt{\pi}}$$

And here I do stop because I cannot go on (mostly because I have no paper and pencils with me.. I'll check again when I'll be at home).


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