# Integrating exponential function raised to a fractional term

I'm trying to integrate an exponential term raised to a fractional power with other variables in it. I'm really rusty and having a hard time trying to figure out where to start. I'd like to pull out the other terms so I can work on integrating but I'm not sure if or how I can do that with it being part of the exponent. Below is the what I am trying to integrate.

$$\int_e{e^{\frac{-x^2\beta^2-y^22\alpha^2}{2\alpha^2\beta^2}}dy}$$

Any help is greatly appreciated.

• what do you meant with $/2\alpha^2\beta^2$ Nov 14, 2015 at 2:46
• Why do you have $\frac{2\alpha^2} {2\alpha^2 \beta^2}$? Strange to have not simplified problem.
– Don
Nov 14, 2015 at 2:58
• Are any of the other variables functions of $y$ (or vice versa)? Nov 14, 2015 at 3:00
• Also,it is just integral with respect to $y$. So you can pull out $e^{-x^2\beta^2}$. So you are left with $\int_e e^{-y^2 \frac{2\alpha^2}{2\alpha^2\beta^2}}dy$=$\int_e e^{-y^2 \frac{1}{\beta^2}}dy$
– Don
Nov 14, 2015 at 3:01

Indefinite integral would be: $$\int{e^{\frac{-x^2\beta^2-y^22\alpha^2}{2\alpha^2\beta^2}}dy}$$=$$\int{e^{\frac{-y^22\alpha^2}{2\alpha^2\beta^2}}e^{\frac{-x^2\beta^2}{2\alpha^2\beta^2}}dy}$$=$$\int{e^{\frac{-y^2}{\beta^2}}e^{\frac{-x^2}{2\alpha^2}}dy}$$=$$e^{\frac{-x^2}{2\alpha^2}}\int{e^{\frac{-y^2}{\beta^2}}dy}$$=$$e^{\frac{-x^2}{2\alpha^2}}(\frac{\sqrt\pi\beta\ erf(\frac{y}{\beta})}{2} +C)$$ So just put the limits and you will get the answer you were looking for.
we ge this here $$1/2\,{{\rm e}^{-1/2\,{\frac {{x}^{2}}{{\beta}^{2}}}}}\sqrt {\pi } \sqrt {2}\alpha\, {{\rm erf}\left(1/2\,{\frac {\sqrt {2}y}{\alpha}}\right)}$$
• I think you switched $\alpha$ with $\beta$
• It would be right if $$\int_e{e^{\frac{-x^2\alpha^2-y^22\beta^2}{2\alpha^2\beta^2}}dy}$$. You switched the constants.