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:
$$\lim_{n\to\infty}n\int_0^1 nx^{n-1}\left(\frac{1}{1+x}-\frac{1}{2}\right)\mathrm dx.$$
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:
$$\lim_{h\rightarrow 0}\int_0^1 \frac{-t^h\ln t}{(1+t^h)^2}\,dt=-\frac{1}{4}\int_0^1\ln t \,dt=\boxed{\dfrac{1}{4}}$$
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