7
$\begingroup$

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

$\endgroup$
8
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
5
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.