Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $0 <a < 1$ and let $b,c \in \mathbb{N}$, evaluate $$\int_0^a \frac{x^{b}+x^{c}}{(1+x)^{b+c+2}}$$

How to evaluate in terms of $a$,$b$ and $c$?

share|improve this question
6  
Without orders please –  no identity May 14 '12 at 12:20
    
can't we take b and c arbitrary? –  dato datuashvili May 14 '12 at 12:26
    
Maple evaluates this in terms of hypergeometric functions. @dato: without the assumption that $b$ is a positive integer, Maple's output has factors like $\csc(\pi b)$ in it. –  GEdgar May 14 '12 at 13:46
    
Great necro-edit... /s –  The Chaz 2.0 Jul 21 '13 at 1:04

2 Answers 2

Write the integral as

$$I(a,b,c)=\int_0^a {{{\left( {\frac{x}{{1 + x}}} \right)}^b}{{\left( {\frac{1}{{1 + x}}} \right)}^c}\frac{{dx}}{{{{\left( {1 + x} \right)}^2}}}} + \int_0^a {{{\left( {\frac{x}{{1 + x}}} \right)}^c}{{\left( {\frac{1}{{1 + x}}} \right)}^b}\frac{{dx}}{{{{\left( {1 + x} \right)}^2}}}} $$

Since $$\frac{x}{{1 + x}} = 1 - \frac{1}{{x + 1}}$$

Let $$u = \frac{1}{{x + 1}}$$

We get

$$ -I(a,b,c)= \int_1^\alpha {{{\left( {1 - u} \right)}^b}{u^c}du} + \int_1^\alpha {{{\left( {1 - u} \right)}^c}{u^b}du} $$

Where $$\alpha = \frac{1}{{a + 1}}$$

Now in any of the two let $u=1-u'$, to get(I'll keep the $u$ variable)

$$-I(a,b,c)=\int_1^\alpha {{{\left( {1 - u} \right)}^b}{u^c}du} - \int_0^{1 - \alpha } {{u^c}{{\left( {1 - u} \right)}^b}du} $$

Now go back to $\alpha = \frac{1}{a+1}$. You should get something like this

$$I\left( {a,b,c} \right) = \int_0^1 {{u^c}{{\left( {1 - u} \right)}^b}du} + \int_{\frac{1}{{a + 1}}}^{\frac{a}{{a + 1}}} {{u^c}{{\left( {1 - u} \right)}^b}du} $$

$$I\left( {a,b,c} \right) = \frac{{\left( {c + 1} \right)!\left( {b + 1} \right)!}}{{\left( {b + c + 2} \right)!}} + \int_{\frac{1}{{a + 1}}}^{\frac{a}{{a + 1}}} {{u^c}{{\left( {1 - u} \right)}^b}du} $$

The last integral might be harder to compute, but given the conditions on $a$, one has

$$0 < \frac{a}{{a + 1}} < \frac{1}{{a + 1}} < 1$$

so the Binomial expansion can be used.

share|improve this answer

Since the question only cares about natural numbers of $b$ and $c$ , it should be no problem about any binomial series expansions.

$\int_0^a\dfrac{x^b+x^c}{(1+x)^{b+c+2}}dx$

$=\int_0^a\dfrac{x^b}{(1+x)^{b+c+2}}dx+\int_0^a\dfrac{x^c}{(1+x)^{b+c+2}}dx$

$=\int_0^a\dfrac{(x+1-1)^b}{(x+1)^{b+c+2}}dx+\int_0^a\dfrac{(x+1-1)^c}{(x+1)^{b+c+2}}dx$

$=\int_0^a\dfrac{\sum\limits_{n=0}^bC_n^b(-1)^n(x+1)^{b-n}}{(x+1)^{b+c+2}}dx+\int_0^a\dfrac{\sum\limits_{n=0}^cC_n^c(-1)^n(x+1)^{c-n}}{(x+1)^{b+c+2}}dx$

$=\int_0^a\sum\limits_{n=0}^bC_n^b(-1)^n(x+1)^{-n-c-2}~dx+\int_0^a\sum\limits_{n=0}^cC_n^c(-1)^n(x+1)^{-n-b-2}~dx$

$=\left[\sum\limits_{n=0}^b\dfrac{C_n^b(-1)^n(x+1)^{-n-c-1}}{-n-c-1}\right]_0^a+\left[\sum\limits_{n=0}^c\dfrac{C_n^c(-1)^n(x+1)^{-n-b-1}}{-n-b-1}\right]_0^a$

$=\sum\limits_{n=0}^b\dfrac{C_n^b(-1)^n}{n+c+1}-\sum\limits_{n=0}^b\dfrac{C_n^b(-1)^n}{(n+c+1)(a+1)^{n+c+1}}+\sum\limits_{n=0}^c\dfrac{C_n^c(-1)^n}{n+b+1}-\sum\limits_{n=0}^c\dfrac{C_n^c(-1)^n}{(n+b+1)(a+1)^{n+b+1}}$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.