Short version:

Can anyone do $$ \int dxe^{-x^{2}}\sqrt{2x+a} $$ ?

Long Version:

In an answer to Integrating a product of exponentials and error functions the integral below is evaluated using a differentiation technique:

Let $\mathcal{I}(\gamma)= \int_0^\infty x^2\exp(-\delta x^2)\operatorname{erf}(\gamma x)\, \mathrm{d} x$, then $I^\prime(\gamma) = \frac{2}{\sqrt{\pi}} \int_0^\infty x^3 \exp(-x^2 \left( \delta + \gamma^2 \right) ) \mathrm{d} x = \frac{1}{(\gamma^2 + \delta)^2} \frac{1}{\sqrt{\pi}}$. Thus: $$ I(\gamma) = \int_0^\gamma \frac{1}{\sqrt{\pi}} \frac{1}{(\gamma^2+\delta)^2} \mathrm{d} \gamma = \frac{\gamma }{2 \sqrt{\pi } \delta \left(\gamma ^2+\delta \right)}+\frac{1}{2 \sqrt{\pi } \delta ^{3/2}} \arctan\left(\frac{\gamma }{\sqrt{\delta }}\right) $$

I have used this method in order to evaluate:

$$ \int_{0}^{\infty}\int_{0}^{\epsilon+\frac{1}{2r}x^{2}}e^{-\beta\left(x^{2}+y^{2}\right)}dydx $$

and my final step involves:

$$ \int dxe^{-x^{2}}\sqrt{2x+a} $$

$a$ is just some constant.

1) Do I need limits on my final integral like in Integrating a product of exponentials and error functions, i.e. $$ \int_{0}^{x}dxe^{-x^{2}}\sqrt{2x+a} $$

2) Can anyone help me evaluate my final integral, whether its indefinite or not? If you can't answer this question completely, does anyone know how to do the final integral $\int dxe^{-x^{2}}\sqrt{2x+a}$?

  • $\begingroup$ Just to be clear, are you looking for a primitive, or do you wish to evaluate that definite-integral on some interval? $\endgroup$ – Jonathan Y. Feb 14 '14 at 23:00
  • 1
    $\begingroup$ hopefully a primitive, though the long version indicates that I might need to integrate over an interval (the referenced question integrates over 0 to gamma). If you can get a primitive that would be good. $\endgroup$ – Alexander Kartun-Giles Feb 14 '14 at 23:08

I doubt there's any kind of closed form primitive in terms of common special functions, but we can try to find a series solution for $t < \infty$:

$$\int_0^t e^{-x^2}\sqrt{x+b}~dx=\sum_{k=0}^\infty \frac{(-1)^k}{k!} \int_0^t x^{2k} \sqrt{x+b}~dx$$

The integral can be transformed to yield a hypergeometric function:

$$\int_0^t x^{2k} \sqrt{x+b}~dx=\sqrt{b} ~t^{2k+1} \int_0^1 y^{2k} \left(1+\frac{t}{b} y\right)^{1/2}~dy= \\ =\frac{\sqrt{b} ~t^{2k+1}}{2k+1} {_2F_1} \left(-\frac{1}{2};2k+1;2k+2;-\frac{t}{b} \right)$$

Which gives us for the original integral:

$$\int_0^t e^{-x^2}\sqrt{x+b}~dx=\sqrt{b} ~t~\sum_{k=0}^\infty \frac{(-1)^k~t^{2k}}{k!(2k+1)}{_2F_1} \left(-\frac{1}{2};2k+1;2k+2;-\frac{t}{b} \right) \tag{1} $$

The hypergeometric function is elementary for integer $k$, but I haven't been able to find a complete closed form. The first few are:

$$\frac{1}{2}~{_2F_1} \left(-\frac{1}{2};1;2;-p \right)=\frac{(1+p)^{3/2}-1}{3p}$$

$$\frac{1}{2}~{_2F_1} \left(-\frac{1}{2};3;4;-p \right)=\frac{(1+p)^{3/2}(8-12p+15p^2)-8}{35p^3}$$

$$\frac{1}{2}~{_2F_1} \left(-\frac{1}{2};5;6;-p \right)=\frac{(1+p)^{3/2}(128-192p+240p^2-280p^3-315p^4)-128}{693p^5}$$

Some of the coefficients can be found in http://oeis.org/A061549 and http://oeis.org/A001803.

Using a simple susbsitution $y=1-z$ we obtain another form of the integral, resulting in another hypergeometric function:

$$\int_0^t e^{-x^2}\sqrt{x+b}~dx=\sqrt{t+b}~ ~t~\sum_{k=0}^\infty \frac{(-1)^k~t^{2k}}{k!(2k+1)}{_2F_1} \left(-\frac{1}{2};1;2k+2;\frac{t}{t+b} \right) \tag{2}$$

Numerically, this series converges slower than the first one. However, by looking at the denominators, it's clear that for exotic case $t=-b$ we should use $(1)$, while for the case $b=0$ we should use $(2)$.

  • $\begingroup$ Yes interesting, thank you. I will try this now. $\endgroup$ – Alexander Kartun-Giles Apr 14 '18 at 11:36

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