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I would like to find out if this integral converges: $$\int_{-\infty}^{\infty} e^{-\sqrt{|x|}}\,\mathrm{d}x$$

Since this is a symmetric function I figured I could focus on only one side of the integral, namely

$\displaystyle\int_{0}^{\infty} e^{-\sqrt{|x|}}\,\mathrm{d}x$ which in this case is equivalent to $\displaystyle\int_{0}^{\infty} e^{-\sqrt{x}}\,\mathrm{d}x$ (since $|x| = x$ when $x > 0$)

Also, $e^{-\sqrt{x}}$ is bounded from 0 to 1 meaning the integral there is a constant, so I will use the integral from 1 to $\infty$.

I know this converges (checked with a calculator) but cannot seem to find an argument for the comparison test to say that since $e^{-\sqrt{x}} < $ "some other function which converges" for $x > 1$, thus $\displaystyle\int_1^{\infty} e^{-\sqrt{x}}\,\mathrm{d}x$ converges.

In other words, I need a function which is always greater than $e^{-\sqrt{x}}$ and whose integral converges. I know that $e^{-x}$ and $e^{-2x}$ both converge, but these are both smaller than $e^{-\sqrt{x}}$ for $x > 1$.

Tips would be appreciated. Thank you.

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  • $\begingroup$ Try with powers, $x^\alpha$. $\endgroup$ – Daniel Fischer Aug 4 '13 at 23:30
  • $\begingroup$ Oops. Added the $\mathrm{d}x$'s. $\endgroup$ – mhy Aug 5 '13 at 19:37
  • $\begingroup$ In general, for positive values of n, we have $\quad\displaystyle\int_0^\infty e^{-\sqrt[n]x}dx=n!$ $\endgroup$ – Lucian Jun 23 '14 at 1:01
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Hint: for $x > 75$, $\ln(x^{2}) < \sqrt{x}$.

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    $\begingroup$ Thanks. I got the answer. To solve these types of problems, how did you come up with $x^2$? I understand the logic - If $e^{\sqrt{x}} > x^2$, then $\sqrt{x} > ln(x^2)$, and thus solving $\sqrt{x} - ln(x^2) = 0$ for x, we get 74.1867... Then this means for $x > 75$ $\sqrt{x} > ln (x^2)$, and $e^{\sqrt{x}} > x^2$, thus $\frac{1}{e^{\sqrt{x}}} < \frac{1}{x^2}$ and we can use the comparison test to say that the integral of the left side converges. $\endgroup$ – mhy Aug 4 '13 at 23:44
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Note that $$\int_0^A e^{-\sqrt x}dx=\int_0^{\sqrt A}2ue^{-u}du$$

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  • $\begingroup$ Thanks. I guess this is a setup for integration by parts? $\endgroup$ – mhy Aug 4 '13 at 23:46
  • $\begingroup$ @mhy Not really. The function $ue^{-u/2}$ is bounded and positive and $e^{-u/2}$ goes to zero exponentially. $\endgroup$ – Pedro Tamaroff Aug 4 '13 at 23:50
  • $\begingroup$ @mhy: of course, if you integrate by parts, you will get the value of $\frac12\int_{-\infty}^\infty e^{-\sqrt{|x|}}\,\mathrm{d}x$ as $A\to\infty$. $\endgroup$ – robjohn Aug 5 '13 at 7:09
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Using L'Hospital, we get $$ \begin{align} \lim_{x\to\infty}xe^{-x} &=\lim_{x\to\infty}\frac{x}{e^x}\\ &=\lim_{x\to\infty}\frac1{e^x}\\[4pt] &=0 \end{align} $$ Therefore, $$ \begin{align} \lim_{u\to\infty}\frac14\sqrt{u}e^{-\frac14\sqrt{u}}&=0&&\text{substitute $x=\frac14\sqrt{u}$}\\ \lim_{u\to\infty}\frac1{256}u^2e^{-\sqrt{u}}&=0&&\text{raise to the $4^{\text{th}}$ power}\\ \lim_{u\to\infty}u^2e^{-\sqrt{u}}&=0&&\text{multiply by $256$}\\ \end{align} $$ Thus, for all $x$, $e^{-\sqrt{|x|}}\le1$, and for $x$ sufficiently large, i.e. $|x|\ge16\,\mathrm{W}_{\!-1\!}\left(-\frac14\right)^2\approx74.186688$, $e^{-\sqrt{|x|}}\le\frac1{x^2}$. This implies that $$ \int_{-\infty}^\infty e^{-\sqrt{|x|}}\,\mathrm{d}x $$ converges.

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