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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:

enter image description hereenter image description here

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?

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    $\begingroup$ +1 for writing effort. But I don't understand what do you mean by analytically solving. The integral of any odd function between symmetric bounds is zero - that's all one needs to say. $\endgroup$ Aug 12, 2013 at 11:53
  • $\begingroup$ I would like to analytically show, that they equal 0. Check my Edited Question. I was thinking to solve them using the relation described there, but i get spapped limits and a negaitve sign... Long story short, I need to know what are the relations between $$\int\limits_{0}^{\infty} dx \qquad \int\limits_{0}^{-\infty} dx \qquad \int\limits_{\infty}^{0} dx \qquad \int\limits_{-\infty}^{0}$$ $\endgroup$
    – 71GA
    Aug 12, 2013 at 12:01
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    $\begingroup$ Hint: once you have decomposed $\int_{-\infty}^{\infty}$ as $\int_{-\infty}^0+\int_0^{\infty}$, look at what happens to the integral $\int_{-\infty}^0$ after the change of variables $x\rightarrow -x$. $\endgroup$ Aug 12, 2013 at 12:01
  • $\begingroup$ I like hints like this one :) $\endgroup$
    – 71GA
    Aug 12, 2013 at 12:04
  • $\begingroup$ I know that $\int_{-\infty}^0 = - \int_{0}^{-\infty}$ but if i want to know what happens if i insert $-x$ instead of $x$ i have to check what function i have. In my case it is odd so i should get the change in sign also... Does this mean that $\int_{-\infty}^{0}=-\int_{\infty}^{0}$ AND $\int_{0}^{\infty}=-\int_{0}^{-\infty}$ ??? $\endgroup$
    – 71GA
    Aug 12, 2013 at 14:17

4 Answers 4

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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.

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  • $\begingroup$ Your case is for the $x$ what about for the $x^2$? $\endgroup$
    – 71GA
    Aug 12, 2013 at 12:03
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    $\begingroup$ Told you: integration by parts $$\int x^2e^{-ax^2}dx=-\frac x{2a}e^{-ax^2}+\frac1a\int xe^{-ax^2}=\ldots$$ $\endgroup$
    – DonAntonio
    Aug 12, 2013 at 12:11
  • $\begingroup$ One more thing. Isn't it $\int e^{ax^2}dx = \frac{1}{2ax}e^{ax^2}$? $\endgroup$
    – 71GA
    Aug 12, 2013 at 12:17
  • $\begingroup$ No. If you multiply by $\,x\,$ the integrand then yes, yet all your exponentials have minus sign in the exponent... $\endgroup$
    – DonAntonio
    Aug 12, 2013 at 12:25
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    $\begingroup$ Because the derivative of $\,\frac1{2ax}e^{ax^2}\;$ is not $\;e^{ax^2}\;$ ...! $\endgroup$
    – DonAntonio
    Aug 12, 2013 at 12:54
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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!)

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  • $\begingroup$ Oh i forgot to mention that $n,~a \in \mathbb{N}$. $\endgroup$
    – 71GA
    Aug 12, 2013 at 12:02
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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".

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Use $x = t^{1/2}$. Then

$$ \int_{0}^{\infty}x^{n}{\rm e}^{-x^{2}}\,{\rm d}x = \int_{0}^{\infty}t^{n/2}{\rm e}^{-t}\,{1 \over 2}\,t^{-1/2}\,{\rm d}t = {1 \over 2}\int_{0}^{\infty}t^{\left(n - 1\right)/2}{\rm e}^{-t}\,{\rm d}t = {1 \over 2}\,\Gamma\left(n + 1 \over 2\right) $$

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