Decide if the following integral is convergent or divergent. If it is convergent evaluate the integral. $$\int_1^{\infty} \frac{e^{-\sqrt {x}}}{\sqrt{x}}\,\mathrm dx$$

I evaluated the integral, but I could not decide via comparison test whether the integral is convergent or divergent.

  • $\begingroup$ Please include your work so we can help you. $\endgroup$ – Michael Albanese Jan 2 '14 at 13:24
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    $\begingroup$ If you evaluated it, wouldn't that be proof enough whether it converges or diverges? $\endgroup$ – Arthur Jan 2 '14 at 13:25
  • $\begingroup$ But it says first decide then evaluate so I tried to compare this function with 1/x^(1/2) , but it is greater and divergent but I couldnt find smaller funstion than given $\endgroup$ – Pumpkin Jan 2 '14 at 13:29
  • $\begingroup$ @Pumpkin For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – user93957 Jan 2 '14 at 14:23

$$e^{\sqrt{x}}=\sum_{n=0}^{\infty}\frac{x^{0.5n}}{n!}$$ $$\implies e^{\sqrt{x}}>x$$ $$\therefore \sqrt{x}e^{\sqrt{x}}>x^{1.5}$$ $$\frac{1}{\sqrt{x}e^{\sqrt{x}}}<x^{-1.5}$$ Since $\int_1^{\infty} x^{-1.5} dx$ is convergent then the $\int_1^{\infty} \frac{1}{\sqrt{x}e^{\sqrt{x}}}$ is convergent.

  • $\begingroup$ The last step should likely read $<x^{-1.5}$ and the integral as well, which would actually make it convergent :-) since $\int_0^\infty x^{1.5} dx = + \infty$... $\endgroup$ – gt6989b Jan 2 '14 at 13:51
  • $\begingroup$ Thanks for the correction. Indeed you're right! $\endgroup$ – John Jan 2 '14 at 13:52
  • $\begingroup$ The integral that John is showing to be convergent does not seem to be the integral that Pumpkin is talking about... $\endgroup$ – ireallydonknow Jan 2 '14 at 15:29
  • $\begingroup$ @ireallydonknow The editor changed the integral incorrectly. $\endgroup$ – John Jan 2 '14 at 15:31

Hint. Try the transformation: $y=\sqrt{x}$.

  • $\begingroup$ How can I apply comparison test with that substitution? $\endgroup$ – Pumpkin Jan 2 '14 at 13:42
  • $\begingroup$ @Pumpkin if you sub correctly your integral becomes $\int e^{-y}$... $\endgroup$ – gt6989b Jan 2 '14 at 13:52
  • $\begingroup$ With the substitution you do not need to apply any test, as you can find the precise value of the integral. $\endgroup$ – Yiorgos S. Smyrlis Jan 2 '14 at 13:52

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