$$\lim\limits_{x \rightarrow \infty} x e^{x^2} \int_x^{\infty}e^{-u^2}du$$

That $x$ on the lower bound of the Gaussian integral is really confusing.

Can we solve it without the fact $\int_{-\infty}^{\infty}e^{-u^2}du = \sqrt\pi \ge \int_x^{\infty}e^{-u^2}du$ ?

(original question have been edited to fix mistakes pointed out in the comments by anon)

  • 1
    $\begingroup$ Do you mean $\int_x^\infty e^{-u^2}du$? You can't have $x$ be the integral's dummy variable and have $x$ appear outside of the integral. You just can't. $\endgroup$ – anon Jul 22 '15 at 5:30
  • $\begingroup$ no but i'll edit to make it clearer. sorry this is exactly how i got it on the exam .... ? why its not possible ? $\endgroup$ – Boris Morozov Jul 22 '15 at 5:30
  • $\begingroup$ Then I'll fix your exam for you - edited. Do you think $\lim\limits_{x\to\infty}x\cdot x^{-1}$ is $\infty$ because $x\to\infty$ but $x^{-1}$ is bounded for $x>1$? Notice how $x^{-1}\to0$ though! $\endgroup$ – anon Jul 22 '15 at 5:37
  • $\begingroup$ i wish you were at the exam to fix it :D your right $\endgroup$ – Boris Morozov Jul 22 '15 at 5:38

Why not rewrite $$x e^{x^2} \int_x^{\infty}e^{-u^2}du=\frac{\int_x^{\infty}e^{-u^2}du}{\frac 1xe^{-x^2}} =\frac u v$$ Use the fundamental theorem of calculus and L'Hospital $$u'=-e^{-x^2}$$ (this comes from the fundamental theorem of calculus) $$v'=-2 e^{-x^2}-\frac{e^{-x^2}}{x^2}$$ $$\frac{u'}{v'}= \frac{-e^{-x^2}}{-2 e^{-x^2}-\frac{e^{-x^2}}{x^2}}=\frac{1}{2-\frac 1 {x^2} }$$

I am sure that you can take from here.


This was the short way. Now, the long long way : assuming that you know the error function $$\int_x^{\infty}e^{-u^2}du=\frac{1}{2} \sqrt{\pi } \,\text{erfc}(x)$$ So $$A=x e^{x^2} \int_x^{\infty}e^{-u^2}du=\frac{1}{2} \sqrt{\pi } e^{x^2} x \, \text{erfc}(x)$$ For large values of the argument, you could find the expansion $$\text{erfc}(x)=e^{-x^2} \left(\frac{1}{\sqrt{\pi } x}-\frac{1}{2 \sqrt{\pi } x^3}+O\left(\left(\frac{1}{x}\right)^4\right)\right)$$ which makes $$A=\frac{1}{2}-\frac{1}{4 x^2}+O\left(\left(\frac{1}{x}\right)^4\right)$$

  • $\begingroup$ You beat me to it! +1 $\endgroup$ – Mark Viola Jul 22 '15 at 6:51

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