How to find the value of this limit? $\lim\limits_{n\to\infty}n\int_0^1 nx^{n-1}\left(\frac{1}{1+x}-\frac{1}{2}\right)\mathrm dx.$ How to calculate the following limit:
$$\lim_{n\to\infty}n\int_0^1 nx^{n-1}\left(\frac{1}{1+x}-\frac{1}{2}\right)\mathrm dx.$$
 A: Use the substitution $x^n=t$.
$$n\int_0^1 nx^{n-1}\left(\frac{1}{1+x}-\frac{1}{2}\right)\,dx=n\int_0^1 \left(\frac{1}{1+t^{1/n}}-\frac{1}{2}\right)\,dt$$
$$\Rightarrow \lim_{n\rightarrow \infty} n\int_0^1 \left(\frac{1}{1+t^{1/n}}-\frac{1}{2}\right)\,dt=\lim_{h\rightarrow 0}\dfrac{\displaystyle \int_0^1 \left(\dfrac{1}{1+t^{h}}-\frac{1}{2}\right)\,dt}{h}$$
Use L'Hopital's rule and Leibniz rule to get:
$$\lim_{h\rightarrow 0}\int_0^1 \frac{-t^h\ln t}{(1+t^h)^2}\,dt=-\frac{1}{4}\int_0^1\ln t \,dt=\boxed{\dfrac{1}{4}}$$
A: My intuition in this situation is to integrate by parts, because we can use it to get rid of the factor of $n$ that we don't like. So use $u=\frac{1}{1+x}-1/2$ and $dv=nx^{n-1} dx$, then the boundary terms will cancel (check this!) and we will be left with
$$n \int_0^1 \frac{x^n}{(1+x)^2} dx$$
Introducing a factor of $n+1$ for convenience, we have
$$\frac{n}{n+1} \int_0^1 \frac{(n+1) x^n}{(1+x)^2} dx$$
We again integrate by parts, with a similar substitution as before. We get
$$\frac{n}{n+1} \left ( 1/4 - \int_0^1 \frac{-2 x^{n+1}}{(1+x)^3} dx \right )$$
(Check the boundary term yourself). It is not hard to show, by comparing to the integral of $2x^{n+1}$, that this last integral goes to zero as $n \to \infty$, and $\frac{n}{n+1} \to 1$ as $n \to \infty$, so the desired limit is $1/4$.
If you are in or have taken real analysis, then I think it is an instructive exercise with uniform convergence to show that if you replace $\frac{1}{1+x}-1/2$ with any continuous function $f$ so that $f(1)=0$ and $f'(1)$ exists, the limit will be $-f'(1)$. (So even though the example looks weird and complicated, it is in a certain sense "natural".) If this is actually for calculus as the tags suggest then please disregard this remark.
A: The integral is equal to (via Mathematica):
$$I(n)=\int_0^1 nx^{n-1}\left(\frac{1}{1+x}-\frac{1}{2}\right)\mathrm dx.=(1/2)\left(-1 - n\psi( n/2) + n\psi((1 + n)/2)\right)$$
Where $\psi(z):=\frac{\Gamma'(z)}{\Gamma(z)}$ is the PolyGamma function.
The limit is equal to (via Mathematica):
$$\lim_{n\to\infty} n I(n)=\frac{1}{4}$$
-mike
