Integrals of form $\int^{∞}_{-∞} x^{2k} \exp(-b^2 (x+x_0)^2)dx$, $k=1,2$.

Question

Please can anybody help me to solve these integrals:

$$\int^{∞}_{-∞} x^2 \exp(-b^2 (x+x_0)^2)dx\,,\,\,\int^{+∞}_{-∞} x^4 \exp(-b^2 (x+x_0)^2)dx\;\;?$$

Answer 1

One may start with the gaussian integral $$ \int^{∞}_{-∞} \exp(-b^2 u^2)du=\frac{\sqrt{\pi}}{b},\qquad b>0, $$ getting, by differentiation with respect to the parameter $b$, $$ \int^{∞}_{-∞} u^2\exp(-b^2 u^2)du=\frac{\sqrt{\pi}}{2b^3}, $$$$ \int^{∞}_{-∞} u^4\exp(-b^2 u^2)dx=\frac{\sqrt{\pi}}{4b^4}. $$ Then, by the change of variable $$ u=x+x_0,\qquad du=dx, $$ one obtains $$\int^{∞}_{-∞} x^2 \exp(-b^2 (x+x_0)^2)dx=\int^{∞}_{-∞} (u-x_0)^2\exp(-b^2 u^2)du $$$$\int^{∞}_{-∞} x^4 \exp(-b^2 (x+x_0)^2)dx=\int^{∞}_{-∞} (u-x_0)^4\exp(-b^2 u^2)du $$ and one may conclude by expanding the integrand, using the parity and the above results.

Answer 2

An idea that you shall complete: substituting in first integral

$$u:=x+x_0\implies du=dx\implies \int_{\Bbb R}x^2e^{-b^2(x+x_0^2)}dx=\int_{\Bbb R}(u-x_0)^2e^{-(bu)^2}du=$$

$$=\int_{\Bbb R}u^2e^{-(bu)^2}du-2x_0\int_{\Bbb R}ue^{-(bu)^2}du+x_0^2\int_{\Bbb R}e^{-(bu)^2}du$$

The first integral above is doable by parts or differentiating under the sign integral, the second one is immediate as $\;u=(-2bu)\cdot\left(-\frac1{2b}\right)\;$, and the last one is a usual Gaussian integral

Answer 3

Hint: Consider the integral $$I(\alpha)=\int_0^\infty x^ne^{-\alpha x^2}dx$$ for $n>0$ and $\alpha>0$, then with substitution $x=\sqrt{\dfrac{u}{\alpha}}$ we have $$I(\alpha)=\dfrac12\alpha^{-\frac{n+1}{2}}\int_0^\infty u^{\frac{n-1}{2}}e^{-u}du=\dfrac12\alpha^{-\frac{n+1}{2}}\Gamma\left(\dfrac{n+1}{2}\right)$$ now for $$J=\int_{-\infty}^\infty x^4e^{-b^2(x+x_0)^2}dx$$ with substitution $x=v-x_0$ then the integral $J$ break up to some integrals like $I(\alpha)$. We must note that for odd integer $n>0$, the integrand $x^ne^{-\alpha x^2}$ is an odd function, so $$\int_{-\infty}^\infty x^ne^{-\alpha x^2}dx=0$$