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Degroot makes a statement in his textbook:

It can be shown that for every positive integer $k$,

$$\int_{-\infty}^{\infty}|x|^ke^{-(x-3)^2}\,dx<\infty$$

Can someone show me how I might prove this statement?

David
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  • Note that your integral is a form of a normal distribution which has all its moments, so for all $k$, $E(X^k)$ when $X\sim N(\mu, \sigma)$ exists, so that integral must exist. – Avraham Jun 18 '14 at 00:51

2 Answers2

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The short answer is "because $|x|^k$ is way smaller than $e^{x^2}$, if $|x|$ is large". If you need more details. . .

For any $k$, the power $x^k$ grows much more slowly than the exponential. More precisely, there exists a positive real number $a$ such that

if $x\ge a$ then $x^k<e^x$.

So if $x\ge a$ we have $$|x|^ke^{-(x-3)^2}\le e^{-x^2+7x-9}\ .$$ If also $x\ge8$ then $-x^2+7x-9<x(7-x)\le-x$ and so $$|x|^ke^{-(x-3)^2}\le e^{-x}\ .$$ Hence $$\int_{\max(a,8)}^\infty |x|^ke^{-(x-3)^2}\,dx\le\int_{\max(a,8)}^\infty e^{-x}\,dx<\infty\ .$$ This takes care of the positive "tail" of the integral, the negative tail is similar, and the bit in between is finite because it is the integral over a closed interval of a continuous function.

David
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    Well, the first sentence needs to be rephrased!! (Maybe you could just drop the negative in the exponential?) – Ted Shifrin Jun 18 '14 at 01:48
  • Or I could have written that sentence first instead of last. . . then I probably would have got it right ;-) – David Jun 18 '14 at 02:19
  • Hmmm... If the integrand goes to zero as x gets large, the integral is not necessarily finite (e.g., $\int_1^{\infty}1/\sqrt x,dx$). So, short answer, no, but your detailed solution is amazing, absolutely amazing. Great answer! – David Jun 18 '14 at 02:43
  • The first sentence in your comment is true: that's why we need to recognise that $|x|^k$ is not just smaller than $e^{x^2}$, but way smaller. – David Jun 18 '14 at 03:22
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For all $x$, $x\lt1+x\le e^x$. Therefore, substituting $x\mapsto x^2/n$ and raising to the power $n$ yields $$ \frac{x^{2n}}{n^n}\le e^{x^2}\implies e^{-x^2}\le\frac{n^n}{x^{2n}}\tag{1} $$ Note that although not proven above, the inequality is true for $n=0$.

Therefore, applying $(1)$ with $n=0$ and $n=k+1$, yields $$ \begin{align} \int_{-\infty}^\infty|x|^ke^{-(x-3)^2}\,\mathrm{d}x &=\int_{-\infty}^\infty|x+3|^ke^{-x^2}\,\mathrm{d}x\\ &\le\int_{-1}^1|x+3|^k\,\mathrm{d}x+\int_{|x|\gt1}|x+3|^k\frac{(k+1)^{k+1}}{x^{2k+2}}\,\mathrm{d}x\tag{2} \end{align} $$ both integrals in $(2)$ are obviously finite.

robjohn
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  • Not sure why the second integral is finite. Here is what I got in Mathematica with $k=20$. Integrate[Abs[x + 3]^k(k + 1)^(k + 1)/x^(2 (k + 2)), {x, 1, [Infinity]}] // N 1.6950110^38 – David Jun 19 '14 at 02:09
  • And that's only to the right of $x=1$. – David Jun 19 '14 at 02:15
  • For $|x|\gt1$, $|x+3|\lt4|x|$, therefore, the integral on the right is less than $$\int_{|x|\gt1}\frac{4^k(k+1)^{k+1}}{|x|^{k+2}}\mathrm{d}x=2^{2k+1}(k+1)^k$$ For $k=20$, this is $6.118\times10^{38}$. This is quite an overestimate since $|x+3|$ is more like $|x|$ when $|x|$ is big. – robjohn Jun 19 '14 at 03:18
  • $|-2|>1$, but $|-2+3|$ is not less than $4(-2)$. – David Jun 19 '14 at 03:28
  • but it is less than $4|-2|$. – robjohn Jun 19 '14 at 04:35
  • I finally figured out your answer, another great answer, but you are right, a huge overestimate. Here is some Mathematica code comparing the actual integral and your bound for $k=6$.

    In[3]:= k = 6

    Out[3]= 6

    In[7]:= Integrate[ Abs[x]^k*Exp[-(x - 3)^2], {x, -[Infinity], [Infinity]}] // N

    Out[7]= 2551.67

    In[8]:= Integrate[Abs[x + 3]^k, {x, -1, 1}] + 2^(2k + 1)(k + 1)^k // N

    Out[8]= 9.63783*10^8

     – David Jun 20 '14 at 18:02