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October 17th, 2014, 11:44 AM  #1 
Newbie Joined: Oct 2014 From: Copenhagen Posts: 1 Thanks: 0  Integral solving / simplification
Hello group, I am having some difficulty in solving the following integral: (observe attachment picture 1) Could you suggest a substitution using the sum of an infinite geometric/Taylor or other progression series which would help in evaluating this mathematical expression? Many thanks in advance 
October 17th, 2014, 12:23 PM  #2 
Senior Member Joined: Oct 2014 From: EU Posts: 224 Thanks: 26 Math Focus: Calculus  If I am not wrong, the indefinite integral is in the form: $\displaystyle \int f(x)^n f'(x) \text{d}x = {f(x)^{n + 1} \over n + 1} +c = F(x)$ Finally, since that one in the picture is an improper integral, you will have: $\displaystyle \lim_{(x \to \infty)} F(x)  \lim_{(x \to 0)} F(x)$ Best wishes. szz PS: use $\displaystyle \LaTeX$ next time, formulas are shown much better. Last edited by szz; October 17th, 2014 at 12:47 PM. 
October 17th, 2014, 12:38 PM  #3 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,257 Thanks: 928 Math Focus: Wibbly wobbly timeywimey stuff.  
October 17th, 2014, 05:24 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,926 Thanks: 2205  
October 18th, 2014, 06:27 AM  #5 
Senior Member Joined: Dec 2013 From: some subspace Posts: 212 Thanks: 72 Math Focus: real analysis, vector analysis, numerical analysis, discrete mathematics 
Hmm... Could it be like this: Let $\displaystyle I = \int_0 ^{\infty} (1  x)^{\beta  1} x^{\alpha  1} \mathrm{d}x$. If one assumes that the integral $\displaystyle I$ does converge, then: Let $\displaystyle x = \frac{1}{1  y}  1 = \frac{y}{1  y}$, and thus $\displaystyle \mathrm{d}x = \frac{\mathrm{d}y}{(1  y)^2}$. This allows one to write the integral: $\displaystyle I = \int_0^1 y^{\alpha  1} (1  y)^{\alpha  \beta} (1  2y)^{\beta  1} \mathrm{d}y$, which can be written in terms of hypergeometric function $\displaystyle _2F_1(a,b;c;x)$: $\displaystyle I = \frac{\Gamma(\alpha) \Gamma(1  \beta  \alpha)}{\Gamma(1  \beta)} \,_2F_1(1\beta,\alpha;1\beta;2)$. After that, it would depend on the values of $\displaystyle \alpha$ and $\displaystyle \beta$ how it simplifies. 
October 18th, 2014, 07:48 AM  #6  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,257 Thanks: 928 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  
October 18th, 2014, 08:13 AM  #7 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra 
In order to converge we require that $$\lim_{x \to \infty}(1 x)^{a_21}x^{a_11} = 0 \implies a_1 + a_2 \lt 2$$and for the limit of the integral to be finite at zero we need$a_1,a_2 \gt 0$.


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