Behaviour of the improper integral $\int_{0}^{1}[1-(1-x)^{n-1}]\frac{dx}{x}$

Question

Consider the integral for $n\geq 2$ $$I(n)=\int_{0}^{1}[1-(1-x)^{n-1}]\frac{dx}{x} $$

Prove that $$\lim_{n\to \infty}\frac{I(n)}{\log n}=1$$

In integral $I(n)$, since $\int_{0}^{1}f(x)dx= \int_0^1 f(1-x) dx$ so we get $$I(n)=\int_{0}^{1}\frac{ 1-x^{n-1} }{1-x} dx $$ So we get by using the sum of $n$ terms of a Geometric Progression $$I(n)=\int_0^1 (1+x+x^2+x^3+...+x^{n-2}) dx$$

So $$I(n)=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{n-1}$$ So we obtain $$I(n)=\sum_{k=1}^{n-1}\frac{1}{k}$$ Now we know that $$\gamma=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac{1}{k} -\log n\right)$$ So we have $$\lim_{n\to \infty}\left(I(n)+\frac{1}{n}-\log n\right)=\gamma$$ I am struggling to prove the asymptotic. Any help will be appreciated.

Answer 1

You're most of the way there. You can write your last equation as $$ 0 = \lim_{n\rightarrow\infty}\left[I(n) +\frac{1}{n} - \ln n -\gamma\right] = \lim_{n\rightarrow\infty}[\ln n]\left[\frac{I(n)}{\ln n} - 1 +\frac{1}{n\ln n} - \frac{\gamma}{\ln n}\right] $$ Since the limit of the first term in brackets is $\infty$ and the limit of the product is $0$, the limit of the second term in brackets must also be $0$. Therefore we have $$ 0 = \lim_{n\rightarrow\infty}\left[\frac{I(n)}{\ln n} - 1 +\frac{1}{n\ln n} - \frac{\gamma}{\ln n}\right] = \lim_{n\rightarrow\infty}\frac{I(n)}{\ln n} - 1 $$ And the result follows.

Answer 2

$$I_n=\int_{0}^{1} \sum_{k=1}^{n-1} (-1)^{k-1}{n-1 \choose k}x^{k-1}=\sum_{k=1}^{n-1}(-1)^{k-1} {n-1\choose k}\frac{1}{k}=\int_{0}^{1} \frac{(1-z)^{n-1}-1}{(1-z-1)}dz$$ $$=\int_{0}^{1} \frac{z^{n-1}-1}{z-1}dz=H_{n-1}$$ Next, we have $$H_{n-1} \sim \gamma+\ln n-1/n.$$ Finally, $$\lim_{n\to \infty} \frac{I_n}{\ln n}=\lim_{n \to \infty} \frac{\gamma+\ln n -1/n}{\ln n}=1.$$