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October 17th, 2014, 11:44 AM   #1
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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
Attached Images
 Integral.JPG (9.5 KB, 0 views)

 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. Thanks from topsquark and DanKor Last edited by szz; October 17th, 2014 at 12:47 PM.
October 17th, 2014, 12:38 PM   #3
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Quote:
 Originally Posted by szz If I am not wrong, the indefinite integral is in the form: $\displaystyle \int f(x)^n f'(x) \text{d}x = {f(x)^{n + 1} \over n + 1} +c = F(x)$
It's a very good thought but doesn't cover the case if a1 and a2 are arbitrary.

@Dankor: Are a1 and a2 arbitrary?

-Dan

October 17th, 2014, 05:24 PM   #4
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Quote:
 Originally Posted by DanKor I am having some difficulty . . .
What makes you think the integral will converge?

 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. Thanks from topsquark and DanKor
October 18th, 2014, 07:48 AM   #6
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Quote:
 Originally Posted by fysmat 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.
Clever! I won't even ask how you figured that out...

-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 $$\lim_{x \to \infty}(1- x)^{a_2-1}x^{a_1-1} = 0 \implies a_1 + a_2 \lt 2$$and for the limit of the integral to be finite at zero we need$a_1,a_2 \gt 0$. Thanks from topsquark and DanKor

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