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October 29th, 2016, 11:05 AM  #1 
Newbie Joined: Oct 2016 From: italy Posts: 9 Thanks: 0  Check if integral converges
Hi guys, I've to verify if $\displaystyle \int_{1}^{+\infty} \frac{e^{x}}{\sqrt x}dx$ converges. I use the convergence criteria: $\displaystyle \frac{e^{x}}{\sqrt x} \leq \frac{1}{\sqrt x}$. The exponent of $\displaystyle x$ is $\displaystyle 1/2$ , for the convergence criteria this integral should not converge, but standing to wolfram (and to the book) it does. What am I doing wrong? 
October 29th, 2016, 11:15 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,117 Thanks: 2369 Math Focus: Mainly analysis and algebra 
The fact that your integrand is less than a divergent integrand tells you nothing. You need to find a divergent integrand that is less than yours to prove divergence or a convergent integrand greater than yours to prove convergence. ($e^{x}$). 
October 29th, 2016, 12:37 PM  #3 
Newbie Joined: Oct 2016 From: italy Posts: 9 Thanks: 0 
So what should I do in this case?

October 29th, 2016, 12:40 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,117 Thanks: 2369 Math Focus: Mainly analysis and algebra 
Try some functions. If your first attempt has failed, try to find another.
Last edited by v8archie; October 29th, 2016 at 01:06 PM. 
October 29th, 2016, 12:59 PM  #5 
Newbie Joined: Oct 2016 From: italy Posts: 9 Thanks: 0 
Wait, a convergent integrand greater than this is $\displaystyle e^{x}$ ! p.s.: Oh you wrote that, didn't notice xD 
October 29th, 2016, 01:06 PM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,117 Thanks: 2369 Math Focus: Mainly analysis and algebra 
Well done.


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