April 28th, 2013, 07:59 AM  #1 
Newbie Joined: Apr 2013 Posts: 11 Thanks: 0  Integrals
Judge if those integrals converge: Thank you. 
April 28th, 2013, 12:48 PM  #2 
Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus  Re: Integrals [color=#000000]I think it is easy to prove that the last two converge, c. It is obvious that we have a problem at point 0, so knowing that close to zero for a small ?, , so the integral converges. d. , so again this integral converges.[/color] 
April 28th, 2013, 01:44 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,641 Thanks: 625  Re: Integrals
1) converges. Transform by y^2 = x and get an integrand 2(y^35)exp(y) 2) diverges. Concatonate pairs of intervals of length ? (starting at ?/2) and you have an integrand ~ 1/x^(2/3) as x > ?. 
April 28th, 2013, 02:03 PM  #4  
Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus  Re: Integrals Quote:
 
April 29th, 2013, 06:29 AM  #5 
Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus  Re: Integrals [color=#000000]It was easier than I expected. b. look at the boxed number, this is the case which converges for p>1. For the first integral you could say that it converges since the expotential in the denominator is always greater that the every power of x and (note that here we have a problem at infinity since that limit at 0 is 0). So you could say that for a very big number M, for all so all the given integrals a,b,c,d converge. [/color] 
April 29th, 2013, 02:11 PM  #6  
Global Moderator Joined: May 2007 Posts: 6,641 Thanks: 625  Re: Integrals Quote:
The usual definition I am familiar with has an integrand of exp(x)/³?x.  
April 29th, 2013, 02:29 PM  #7  
Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus  Re: Integrals Quote:
 
April 30th, 2013, 09:37 AM  #8  
Math Team Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions  Re: Integrals Quote:
 
May 4th, 2013, 01:05 PM  #9  
Math Team Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions  Re: Integrals Quote:
 
May 4th, 2013, 02:14 PM  #10  
Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus  Re: Integrals Quote:
See here. I didn't respond earlier because I needed time to do the computations. I used mathematica to get the result I posted earlier, so I didn't have an answer by that time.[/color]  

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