# Showing $(-1)^{n-1}{m\over n+m}\int_{0}^{1}x^{n-1}(1-x)^{m-1}dx=\sum_{i=0}^{n}(-1)^i{n\choose i}H_{n+m-i}$

Showing

$$I=(-1)^{n-1}{m\over n+m}\int_{0}^{1}x^{n-1}(1-x)^{m-1}dx=\sum_{i=0}^{n}(-1)^i{n\choose i}H_{n+m-i}\tag1$$ Where $H_n$ is the nth harmonic number

Recall

$$H_n=\int_{0}^{1}{1-x^n\over 1-x}dx\tag2$$

Sub $(2)$ into $(1)\rightarrow (3)$

$$I=\sum_{i=0}^{n}(-1)^i{n\choose i}\int_{0}^{1}{1-x^{n+m-i}\over 1-x}dx\tag3$$

Let $$J=\int_{0}^{1}{1-x^{n+m-i}\over 1-x}dx\tag4$$

$$J=\sum_{k=0}^{\infty}\int_{0}^{1}{x^k-x^{k+n+m-i}}dx\tag5$$

$$J=\sum_{k=0}^{\infty}\left({1\over k+1}-{1\over k+1+n+m-i}\right)\tag6$$

Can someone help me here to prove I, don't seem to have a clue where I am going?

• Use $H_n=\int\limits_0^1 \frac{1-(1-x)^n}{x}dx$ instead of (2). – user90369 Jun 9 '16 at 13:17
• Out of curiosity, what is the source of all these problems? – Jack D'Aurizio Jun 9 '16 at 13:50
• You mean where did I get them from or what is the point of me posting these problems? – gymbvghjkgkjkhgfkl Jun 9 '16 at 14:24
• @China cat: Have you solved the problem with the hint above ? It's not difficult. – user90369 Jun 9 '16 at 17:02

I think that you can not use this approach because $${1\over k+1}-{1\over k+1+n+m-i}={1\over k+1}-{1\over k+2}+{1\over k+2}-{1\over k+3}+...+{1\over k+1+n+m-i-1}-{1\over k+1+n+m-i}$$ $$J=1+\frac{1}{2}+\frac{1}{3}+...+{1\over 1+n+m-i-1}= \sum\limits_{j=1}^{1+m+n-i-1}{\frac{1}{j}}$$