Proving that $I=\int_0^\infty \frac{n x^{n-1}}{(1+x)^{n+1}}dx<+\infty$

Question

I got this integral from a probability question in which I am interested to prove that it is $<\infty$ $$I=\int_0^\infty \frac{n x^{n-1}}{(1+x)^{n+1}}dx,~~~n \ge 1$$

I tried substitution $y =x+1$, but it does not help, so I am thinking maybe a direct integration is not the way to go. I tried bounding the integrand using $\frac{1}{(1+x)^{n+1}}<1 $ but I get $+\infty$ Any idea, how to do it?

Answer 1

Hint $$\int_0^\infty \frac{n x^{n - 1}}{(x + 1)^{n + 1}} \,dx \leq n \int_0^\infty \frac{(x + 1)^{n - 1}}{(x + 1)^{n + 1}} \,dx .$$ N.b. the inequality requires the condition $n - 1 \geq 0$, since only for those values is $x \mapsto x^{n - 1}$ a nondecreasing function.

Answer 2

$$I=\int_0^\infty \frac{n x^{n-1}}{(1+x)^{n+1}}dx=n\cdot B(n,1)=\frac{n\cdot \Gamma(n)}{\Gamma(n+1)}=1$$

Answer 3

The integrand has a direct antiderivative, which we can see by rewriting the terms like so

$$I = \int_0^\infty n\left(\frac{x}{x+1}\right)^{n-1}\frac{dx}{(x+1)^2} = \int_0^\infty n\left(1-\frac{1}{x+1}\right)^{n-1} d\left(1-\frac{1}{x+1}\right)$$ $$ = \left(1-\frac{1}{x+1}\right)^n\Biggr|_0^\infty = 1$$

Answer 4

We have that $\frac{n x^{n-1}}{(1+x)^{n+1}} \sim \frac n{x^2}$ and this guarantees convergence.

More in detail since

$$\frac{\frac{n x^{n-1}}{(1+x)^{n+1}}}{\frac1{x^2+1}}=\frac{n x^{n+1}+x^{n-1}}{(1+x)^{n+1}}\to n$$

the given integral $I$ converges by limit comparison test with $\int_0^\infty \frac1{x^2+1}dx$.

Answer 5

Splitting the integral into two, we have $$ \int_0^{\infty} \frac{n x^{n-1}}{(1+x)^{n+1}} d x =\int_0^{1} \frac{n x^{n-1}}{(1+x)^{n+1}} d x+ \int_1^{\infty} \frac{n x^{n-1}}{(1+x)^{n+1}} d x $$ The first integral is finite, we need only to consider the second one by letting $x\mapsto\frac{1}{x} $.

$$ \begin{aligned} \int_1^{\infty} \frac{n x^{n-1}}{(1+x)^{n+1}} d x& =\int_0^1 \frac{n}{(1+x)^{n+1}} d x \\ & =\left[-\frac{1}{(1+x)^n}\right]_0^1 \\ & =-\frac{1}{2^n}+1 \\ & <1 \end{aligned} $$ So the integral is finite.

Answer 6

$$\int \frac{n x^{n-1}}{(1+x)^{n+1}}\,dx=\frac {x^n}{(1+x)^n}$$ $$\int_0^p \frac{n x^{n-1}}{(1+x)^{n+1}}\,dx=\left(\frac{p}{p+1}\right)^n=1-\frac{n}{p}+\frac{n (n+1)}{2 p^2}+O\left(\frac{1}{p^3}\right)$$