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

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

How to review in regards to $a$, $b$ and also $c$?

4
2022-07-25 20:41:58
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Answers: 1

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.

4
2022-07-25 22:29:12
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