Prove $ \int_{-\infty}^{\infty} e^{-\alpha(x-c)^2} x^{2n} dx =\int_{-\infty}^{\infty} e^{-\alpha x^2} x^{2n} dx$

Question

I understand
$$ \int_{-\infty}^{\infty} e^{-\alpha x^2} x^{2n} dx =\frac{(2n-1)!!}{2^n}\sqrt{\frac{\pi}{\alpha^{2n+1}}} \quad (\alpha \in \mathbb{C}, \, \text{Re} \,\alpha>0) $$

But I can't prove the following generalized version. $$ \int_{-\infty}^{\infty} e^{-\alpha(x-c)^2} x^{2n} dx =\int_{-\infty}^{\infty} e^{-\alpha x^2} x^{2n} dx \quad (c \in \mathbb{C}, \,\alpha \in \mathbb{C}, \, \text{Re} \,\alpha>0) $$
How can I prove?

I understand well if it goes
$$ \int_{-\infty}^{\infty} e^{-\alpha(x-c)^2} (x-c)^{2n} dx =\int_{-\infty}^{\infty} e^{-\alpha x^2} x^{2n} dx \quad (c \in \mathbb{C}, \,\alpha \in \mathbb{C}, \, \text{Re} \,\alpha>0) $$

Answer 1

This is not an answer but it is too long for a comment.

Changing variable $x=y+c$, it is obvious that $$I_n=\int_{-\infty}^{\infty} e^{-\alpha(x-c)^2} x^{2n} dx =\int_{-\infty}^{\infty} e^{-\alpha y^2} (y+c)^{2n} dx$$ but $$I_n \neq \int_{-\infty}^{\infty} e^{-\alpha y^2} y^{2n} dy$$ which, as @user1952009 commented, would not depend on $c$.

I computed a few of these integrals $$I_1=\frac{\sqrt{\pi } \left(2 \alpha c^2+1\right)}{2 \alpha ^{3/2}}$$ $$I_2=\frac{\sqrt{\pi } \left(4 \alpha ^2 c^4+12 \alpha c^2+3\right)}{4 \alpha ^{5/2}}$$ $$I_3=\frac{\sqrt{\pi } \left(8 \alpha ^3 c^6+60 \alpha ^2 c^4+90 \alpha c^2+15\right)}{8 \alpha ^{7/2}}$$ $$I_4=\frac{\sqrt{\pi } \left(16 \alpha ^4 c^8+224 \alpha ^3 c^6+840 \alpha ^2 c^4+840 \alpha c^2+105\right)}{16 \alpha ^{9/2}}$$ To compute these integrals, you need to expand $(y+c)^{2n}$ using the binomial theorem $$(y+c)^{2n}=\sum_{k=0}^{2n}\binom{2 n}{k}y^k c^{2n-k} $$ and notice that, for odd powers of $k$, $$\int_{-\infty}^{\infty} e^{-\alpha y^2} y^{2k+1} dy=0$$ while $$\int_{-\infty}^{\infty} e^{-\alpha y^2} y^{2k} dy=\alpha ^{-(k+\frac{1}{2})} \Gamma \left(k+\frac{1}{2}\right)$$

I suppose that what they wanted is to show that

$$I_n=\int_{-\infty}^{\infty} e^{-\alpha(x-c)^2} x^{2n} dx =\sum_{k=0}^{n} C_{2k}\int_{-\infty}^{\infty} e^{-\alpha y^2} y^{2k} dx$$

Answer 2

Some notes:

$$F_n(c) = \int_{-\infty}^\infty e^{-(x-c)^2} x^{2n}\, dx= \int_{-\infty}^\infty e^{-x^2} (x+c)^{2n}\, dx$$

• $F_0(c)$ is actually constant as $c$ varies
• For other positive integer $c$, $F_n(c)$ is not constant as $c$ varies. One way of seeing this is to differentiate with respect to $c$ $2n$ times:

$$F^{(2n)}_n(c)=\frac{d^{2n}}{dc^{2n}}\int_{-\infty}^\infty e^{-x^2} (x+c)^{2n}\, dx=\int_{-\infty}^\infty e^{-x^2}\frac{d^{2n}}{dc^{2n}} (x+c)^{2n}\, dx=\int_{-\infty}^\infty e^{-x^2}(2n)!\,dx\neq0$$

As a derivative of $F_n$ is not identically zero, $F_n$ itself must not be constant.