# Exponential integral with square root in power

I have been trying to solve an integral. I know that the solution exists for the form

$$1- \dfrac{2}{\mathcal{R}^2}\int_{0}^{\mathcal{R}_{\mathcal{G}}} \exp (-\Phi r^{\alpha}) r\, {\rm d}r,$$

where $$\alpha>0, \mathcal{R}>0$$. However, I want to solve

$$1- \dfrac{2}{\mathcal{R}^2}\int_{0}^{\mathcal{R}} \exp \left(-\Phi \left(\sqrt{r^2+h^2}\right)^{\alpha}\right) r\, {\rm d}r,$$

where $$\Phi>0$$, $$\alpha>0, \mathcal{R}>0$$. I have tried, but on substituting $$t=r^2+h^2$$, the lower limit of integral becomes h and I checked Table of Integrals (3.351) for an equivalent solution but no hope. Any help on this please?

• Change variable ordinally, and then use polar coordinates. Commented Sep 21, 2018 at 3:29

Using expansion of $$e^x$$ we see $$1- \dfrac{2}{\mathcal{R}^2}\int_{0}^{\mathcal{R}} \exp \left(-\Phi \left(\sqrt{r^2+h^2}\right)^{\alpha}\right) r\, {\rm d}r,$$ $$1- \dfrac{2}{\mathcal{R}^2}\int_{0}^{\mathcal{R}} \sum_{n\geq0}\dfrac{1}{n!}\left(-\Phi \left(\sqrt{r^2+h^2}\right)^{\alpha}\right)^n r\, {\rm d}r,$$ $$1- \dfrac{2}{\mathcal{R}^2} \sum_{n\geq0}\dfrac{(-\Phi )^n}{2n!}\int_{0}^{\mathcal{R}} \left(r^2+h^2\right)^{n\alpha/2} 2r\, {\rm d}r,$$ $$1- \dfrac{2}{\mathcal{R}^2} \sum_{n\geq0}\dfrac{(-\Phi )^n}{2n!(\frac{n\alpha}{2}+1)}\left[ \left(\mathcal{R}^2+h^2\right)^{\frac{n\alpha}{2}+1} -\left(h\right)^{n\alpha+2}\right]$$

Doing the same as Dinesh Shankar in his/her answer. $$f(R)=1- \dfrac{2}{{R}^2}\int_{0}^R \exp \left(-\Phi \left(\sqrt{r^2+h^2}\right)^{\alpha}\right)\, r\, dr$$ $$f(R)=1-\frac{2 }{\alpha\, R^2\,\Phi ^{2/\alpha }}\left( \Gamma \left(\frac{2}{\alpha },\left(h^2\right)^{\alpha /2} \Phi \right)-\Gamma \left(\frac{2}{\alpha },\left(h^2+R^2\right)^{\alpha /2} \Phi \right)\right)$$

• I'm a man. I should have used the OP notation. :) Commented Sep 23, 2018 at 18:23

Your integral is something like that

$$I=\int x\, \mathrm{e}^{-a\left(x^2+b^2\right)^\frac{n}{2}}\,\mathrm{d}x$$

The change of variable $$u=a^\frac{2}{n}\left(x^2+b^2\right)$$ give us

$$I=\frac{1}{2a^\frac{2}{n}}\int\mathrm{e}^{-u^\frac{n}{2}}\,\mathrm{d}u$$

The last integral is related to the incomplete Gamma function:

$$\int\mathrm{e}^{-u^\frac{n}{2}}\,\mathrm{d}u= -\frac{2\operatorname{\Gamma}\left(\frac{2}{n},u^\frac{n}{2}\right)}{n}.$$

So,

$$I= -\frac{\operatorname{\Gamma}\left(\frac{2}{n},u^\frac{n}{2}\right)}{na^\frac{2}{n}}=-\frac{\operatorname{\Gamma}\left(\frac{2}{n},a\left(x^2+b^2\right)^\frac{n}{2}\right)}{na^\frac{2}{n}}.$$

• I've checked it numerically for $a=b=1$, and $n=3$, and $x\in [0,3]$. Commented Sep 21, 2018 at 3:19