Firstly, I know that the Gaussian Integral formula, e.g., $\int^{+\infty}_{-\infty}e^{-ax^2}dx=\sqrt{\frac{\pi}{a}}$. But, I am now being encountered a problem when the integral region is not $[-\infty,+\infty]$. Specifically, I would like to calculate the integral as $\int^b_ce^{-ax^2}dx$. Is there any possible solution?
Besides, I would like to further calculate integral as $\int^b_cxe^{-ax^2}dx$ and $\int^b_cx^2e^{-ax^2}dx$. Is there any trick that I can follow? If you can show some specific deduction or give some references for me, I will highly appreciate your help and time. Thanks for your time and help in advance.
Here are some results : the first come from the definition of the $erf$ function, the other are obtained by integration by parts.
$$\int e^{-ax^2}dx=\frac{\sqrt{\pi } \text{erf}\left(\sqrt{a} x\right)}{2 \sqrt{a}}$$ $$\int x e^{-ax^2}dx=-\frac{e^{-a x^2}}{2 a}$$ $$\int x^2 e^{-ax^2}dx=\frac{\sqrt{\pi } \text{erf}\left(\sqrt{a} x\right)}{4 a^{3/2}}-\frac{x e^{-a x^2}}{2 a}$$ $$\int x^3 e^{-ax^2}dx=-\frac{e^{-a x^2} \left(a x^2+1\right)}{2 a^2}$$ $$\int x^4 e^{-ax^2}dx=\frac{3 \sqrt{\pi } \text{erf}\left(\sqrt{a} x\right)}{8 a^{5/2}}-\frac{x e^{-a x^2} \left(2 a x^2+3\right)}{4 a^2}$$ There is a generalization using the gamma function $$\int x^k e^{-ax^2}dx=-\frac{1}{2} x^{k+1} \left(a x^2\right)^{\frac{1}{2} (-k-1)} \Gamma \left(\frac{k+1}{2},a x^2\right)$$
Edit (8 years later)
Let $ax^2=t$ to make
$$\int x^k e^{-ax^2}\,dx=\frac{1}{2} a^{-\frac{k+1}{2}}\int t^{\frac{k-1}{2}}\,e^{-t}\,dt=-\frac{1}{2} a^{-\frac{k+1}{2}}\, \Gamma \left(\frac{k+1}{2},t\right)$$