# Deciding whether this integral is convergent or divergent: $\int_1^\infty \frac{e^{\sqrt {x}}}{\sqrt{x}}\,\mathrm dx$

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.

• Please include your work so we can help you. Commented Jan 2, 2014 at 13:24
• If you evaluated it, wouldn't that be proof enough whether it converges or diverges? Commented Jan 2, 2014 at 13:25
• 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 Commented Jan 2, 2014 at 13:29
• @Pumpkin For some basic information about writing math at this site see e.g. here, here, here and here.
– user93957
Commented Jan 2, 2014 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.

• The last step should likely read $<x^{-1.5}$ and the integral as well, which would actually make it convergent :-) since $\int_0^\infty x^{1.5} dx = + \infty$... Commented Jan 2, 2014 at 13:51
• Thanks for the correction. Indeed you're right!
– John
Commented Jan 2, 2014 at 13:52
• The integral that John is showing to be convergent does not seem to be the integral that Pumpkin is talking about... Commented Jan 2, 2014 at 15:29
• @ireallydonknow The editor changed the integral incorrectly.
– John
Commented Jan 2, 2014 at 15:31

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

• How can I apply comparison test with that substitution? Commented Jan 2, 2014 at 13:42
• @Pumpkin if you sub correctly your integral becomes $\int e^{-y}$... Commented Jan 2, 2014 at 13:52
• With the substitution you do not need to apply any test, as you can find the precise value of the integral. Commented Jan 2, 2014 at 13:52

Let $$u=\sqrt{x}$$ Then, $$2du=\frac{dx}{\sqrt{x}}$$. $$\sqrt{1} = 1$$, and $$\sqrt{\infty}$$ should be $$\infty$$ when we take limits. So, the integral becomes $$2\int_1^{\infty} e^{-u} du = -2e^{-u}|_1^{\infty} = \frac{2}{e}$$, again, when we use limits. Thus, the integral converges as our value is finite.