# Calculate the Definite Integral.

I need to calculate the integral of a function $$f(x)=x^2e^{-x^2}$$ in the range of $a\leq |x|\leq b$.

My attempt: The integral can be divided in two integral according to range $$I=\int_{-b}^{-a}f(x)dx + \int_a^bf(x)dx$$ As it is clear that $f(x)$ is even function, so $$I=2\int_a^bf(x)dx$$ Now, integral can be divided into $$\int_{a}^{b}f(x)dx = \int_0^bf(x)dx-\int_0^af(x)dx$$Now by using this definition Integral, we can solve the Integral $$\int_0^ux^2e^{-q^2x^2}=\frac{1}{2q^3}\left(\frac{\sqrt{\pi}}{2}\Phi(qu)-qu e^{-q^2u^2} \right)$$ where $\Phi$ is CDF of standard normal distribution.

I am not sure whether I am correct or not.

• What is $q$? In general, this integral can be computed via integration by parts. – Oles Wohnzimmer Dec 14 '16 at 10:33
• Write the integrand as $x.xe^{-x^{2}}$ and then use parts on $u=x, v=xe^{-x^{2}}$ – Kevin Dec 14 '16 at 10:35
• @ Oles Wohnzimmer: $q$ is a constant. – Marcus Dec 14 '16 at 10:43
• I am more curious about limits. Is my attempt of handling with limits is wrong?? – Marcus Dec 14 '16 at 10:44
• What is $\Phi$? – 5xum Dec 14 '16 at 10:45

$$\int x^{2} \mathrm{e}^{-x^{2}} dx = \frac{1}{2} \sqrt{\pi} x^{2} \mathrm{erf}(x) - \sqrt{\pi} \int x \, \mathrm{erf}(x) dx$$
Thus we have $$\int x^{2} \mathrm{e}^{-x^{2}} dx = \frac{\sqrt{\pi}}{4} \mathrm{erf}(x) - \frac{1}{2} x \mathrm{e}^{-x^{2}}$$