User Name Remember Me? Password

 Calculus Calculus Math Forum

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
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
Math Team

Joined: May 2013
From: The Astral plane

Posts: 2,257
Thanks: 928

Math Focus: Wibbly wobbly timey-wimey stuff.
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
Global Moderator

Joined: Dec 2006

Posts: 20,926
Thanks: 2205

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
Math Team

Joined: May 2013
From: The Astral plane

Posts: 2,257
Thanks: 928

Math Focus: Wibbly wobbly timey-wimey stuff.
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 Tags integral, integrals, integration, simplification, solving Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post on4now4 Calculus 2 June 29th, 2014 12:11 PM rsashwinkumar Calculus 0 January 14th, 2013 01:42 AM xerophyte Calculus 5 December 24th, 2010 11:38 PM honzik Real Analysis 1 December 26th, 2009 03:16 PM nimigi Complex Analysis 0 February 28th, 2009 12:30 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.      