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From numerical evidence it appears that whenever the integral converges, $$J_a :=\int_0^1 dx \frac{1+x^a}{(1+x)^{a+2}} = \frac{1}{a+1}.$$ For $a \in \mathbb{N}$, I was able to prove this using induction (see below). How can we prove it for non-integer $a$?

Integrating by parts, $$\begin{align} J_n &= \left.-\frac{1}{n+1} \frac{1+x^n}{(1+x)^{n+1}}\right\lvert_0^1+ \frac{n}{n+1} \int_0^1 dx \frac{x^{n-1}}{(1+x)^{n+1}} \\&=\frac{1}{n+1}\left(1-2^{-n} \right) + \frac{n}{n+1} \int_0^1 dx \left[ \frac{1+x^{n-1}}{(1+x)^{n+1}} - \frac{1}{(1+x)^{n+1}} \right] \\&=\frac{n}{n+1}J_{n-1} + \frac{1}{n+1} \left[\left(1-2^{-n} \right) -\int_0^1 dx \frac{n}{(1+x)^{n+1}} \right] \\&=\frac{n}{n+1}J_{n-1}. \end{align}. $$ Because $J_0 = 1$, we therefore have $$J_n = \frac{1}{n+1} \mathrm{for\,} n \in \mathbb{N}.$$

user111187
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2 Answers2

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$$\begin{align*} \int_{0}^{1}\frac{1+x^{a}}{\left(1+x\right)^{a+2}}dx &=\int_{0}^{1}\frac{1}{\left(1+x\right)^{a+2}}dx+\int_{0}^{1}\frac{x^{a}}{\left(1+x\right)^{a+2}}dx \\ &=\left.-\frac{1+x}{\left(a+1\right)\left(1+x\right)^{a+2}}\right|_{0}^{1}+\left.\frac{\left(1+x\right)x^{a+1}}{\left(a+1\right)\left(1+x\right)^{a+2}}\right|_{0}^{1} \\ &=\frac{1}{a+1}. \end{align*}$$

alexwlchan
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Marco Cantarini
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In general, an integral representation of the beta function is $$B(x,y) = \int_{0}^{1} \frac{t^{x-1} + t^{y-1}}{(1+t)^{x+y}} \ dt \ , \ \text{Re} (x), \text{Re}(y) >0 . $$

This can be derived by expressing the more well-known integral representation $$ B(x,y) = \int_{0}^{\infty}\frac{t^{x-1}}{(1+t)^{x+y}} \ dt \tag{1}$$ as

$$\begin{align} B(x,y) &= \frac{1}{2} \int_{0}^{\infty} \frac{t^{x-1}+t^{y-1}}{(1+t)^{x+y}} \ dt \\ &= \frac{1}{2} \int_{0}^{1} \frac{t^{x-1}+t^{y-1}}{(1+t)^{x+y}} \ dt + \frac{1}{2} \int_{1}^{\infty} \frac{t^{x-1}+t^{y-1}}{(1+t)^{x+y}} \ dt \end{align}$$

and then making the substitution $t = \frac{1}{u}$ in the second integral to get that

$$ \begin{align} B(x,y) &= \frac{1}{2} \int_{0}^{1} \frac{t^{x-1}+t^{y-1}}{(1+t)^{x+y}} \ dt + \frac{1}{2} \int_{0}^{1} \frac{u^{y-1}+u^{x-1}}{(u+1)^{x+y}} \ du \\ &= \int_{0}^{1} \frac{t^{x-1}+t^{y-1}}{(1+t)^{x+y}} \ dt. \end{align}$$

Therefore,

$$ \int_{0}^{1} \frac{1+x^{a}}{(1+x)^{a+2}} \ dx = B(1,a+1) = \frac{\Gamma(a+1)}{\Gamma(a+2)} = \frac{1}{a+1}.$$

$ $

$(1)$ http://mathworld.wolfram.com/BetaFunction.html (22)

Random Variable
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