Some exponential integrals - I need algebraical solution besides my graphical one I have come across integrals of form:
\begin{align}
&\int\limits_{-\infty}^{+\infty} x\cdot e^{-ax^2} dx\\
&\int\limits_{-\infty}^{+\infty} x^2\cdot e^{-ax^2} dx\\
&\int\limits_{-\infty}^{+\infty} x^3\cdot e^{-ax^2} dx\\
&\int\limits_{-\infty}^{+\infty} x^4\cdot e^{-ax^2} dx\\
\end{align}
Where I have figured out after ploting them that for the ones that have the even exponent ($x^2$, $x^4\dots$) I can write the integral like this: 
\begin{align}
&2\int\limits_{0}^{+\infty} x^2\cdot e^{-ax^2} dx\\
&2\int\limits_{0}^{+\infty} x^4\cdot e^{-ax^2} dx\\
\end{align}
I have found these integrals in the Bronštein-Semendijajev mathematics manual [page 474] where he states that we can solve them using the formula:
\begin{align}
\int\limits_{0}^{\infty}x^n \cdot e^{-ax^2}dx = \frac{1\cdot3\dots(2k-1)\,\,\sqrt{\pi}}{2^{k+1}a^{k+1/2}}\longleftarrow\substack{\text{$n$ is the exponent over $x$}\\\text{while $k=n/2$}}
\end{align}
Ok so I can solve these with no problem. But there remains the ones with odd exponent ($x$, $x^3\dots$). On the same page there is a formula for odd exponents, which has a solution:
\begin{align}
\int\limits_{0}^{\infty}x^n \cdot e^{-ax^2}dx = \frac{k\text{!}}{2a^{k+1}}\longleftarrow\substack{\text{$n$ is the exponent over $x$}\\\text{while $k=n/2$}}
\end{align}
but in my case I have odd functions and I cannot use the relation: 
$$\int\limits_{-\infty}^{\infty}dx = 2\int\limits_{0}^{\infty}dx$$
This is why I can't get the form which the mathematical manual needs. When I plotted these even functions I got plots like for example:

From the images I can clearly see that definite integrals between limits $-\infty$ and $\infty$ will equal $0$ for the odd functions. 
Question:
Graphical solution for the integrals odd functions looks easy while I can't seem to use my mathematics manual to solve them analytically. I am wondering if there is analytical way to show that they equal zero. I was thinking about using relation:
$$\int\limits_{-\infty}^{\infty}dx=\int\limits_{-\infty}^{0}dx + \int\limits_{0}^{\infty}dx$$ 
somehow. This way I would get similar form that the manual needs, but with swapped integration limits and sign... How do I solve theese?
 A: Let $n \in \mathbb{N}$ an odd number and consider the integral $\int_{\mathbb{R}} x^{n} e^{-ax^{2}} \: dx$. Using the change of variables $t=-x$ ($dt=-dx$), you get :
$$ \int_{\mathbb{R}} x^{n} e^{-ax^{2}} \: dx = - \int_{\mathbb{R}} t^{n} e^{-at^{2}} \: dt$$
So, $\int_{\mathbb{R}} x^{n} e^{-ax^{2}} \: dx = 0$. (I hope I got your question right!)
A: For example
$$\int\limits_{-\infty}^\infty xe^{-ax^2}dx=-\frac1{2a}\int\limits_{-\infty}^\infty(2ax\,dx)e^{-ax^2}=\left.-\frac1{2a}e^{-ax^2}\right|_{-\infty}^\infty=0$$
All the rest follow from integrating by parts and/or a little inductive argument.
A: For odd $n$,
letting $y = x^2$
so $dy = 2x\ dx$,
$\int x^{2m+1}e^{-x^2} dx
=(1/2)\int y^m e^{-y} dy
$
and this can be done explicitly
by repeated integration by parts
to get
$\int y^m e^{-y} dy
=m!\left(1-e^{-y}\sum_{k=0}^m \dfrac{y^k}{k!}\right)
$.
For more general $n$,
look up "incomplete Gamma function".
A: Use $x = t^{1/2}$. Then
$$
\int_{0}^{\infty}x^{n}{\rm e}^{-x^{2}}\,{\rm d}x
=
\int_{0}^{\infty}t^{n/2}{\rm e}^{-t}\,{1 \over 2}\,t^{-1/2}\,{\rm d}t
=
{1 \over 2}\int_{0}^{\infty}t^{\left(n - 1\right)/2}{\rm e}^{-t}\,{\rm d}t
=
{1 \over 2}\,\Gamma\left(n + 1 \over 2\right)
$$
