# Evaluating $\int_0^a \frac{x^{b}+x^{c}}{(1+x)^{b+c+2}}$

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 $c$?

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Without orders please – Norbert 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

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.

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

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

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

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

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

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

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

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

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