What will be the integral of this natural log function?

Question

I want to do this integral: $$\int_{0}^{1} (-L) \gamma_{avg} ln(1-p) p^L dp$$ the solution provided is : $$\gamma_{avg} \sum_{k=1}^{L} \frac{1}{k} $$ But I am not able to get it, I even tried integration with parts, but of no use.

Answer 1

Let $$A(L)=(1+L)\int_0^1 p^L\ln(1-p)dp$$

Then, using integration by parts: $$\int_0^1\ln(1-p)p^Ldp=$$ $$\underbrace{\bigg[p^L(p-1)\ln(1-p)-p^{L+1}\bigg]_0^1}_{-1}-L\int_0^1 p^{L-1}(p-1)\ln(1-p)dp+\underbrace{L\int_0^1 p^Ldp}_{\dfrac L{L+1}}$$ $$\int_0^1\ln(1-p)p^Ldp=-L\int_0^1 p^L\ln(1-p)dp+L\int_0^1 p^{L-1}\ln(1-p)-\frac 1{L+1}$$ $$\color{red}{\implies}(1+L)\int_0^1\ln(1-p)p^Ldp=L\int_0^1 p^{L-1}\ln(1-p)dp-\frac 1{L+1}$$ $$\color{red}{\implies} A(L)=A(L-1)-\frac 1{L+1}$$ $$A(L)=A(L-2)-\frac 1L-\frac1{L+1}$$ $$\cdots$$ $$A(L)=A(0)-\sum_{k=2}^{L+1} \frac 1k$$ $$A(0)=\int_0^1 \ln(1-p)dp=-1$$ $$A(L)=-\sum_{k=1}^{L+1} \frac 1k$$ $$\int _0^1 p^L\ln(1-p)dp=-\frac 1{L+1}\sum_{k=1}^{L+1} \frac 1k$$ If in fact what you are really looking for is: $$\int_0^1 (-L)\gamma \ln(1-p)p^{\color{red}{L-1}}dp$$ then $$A(L-1)=L\int_0^1 \ln(1-p)p^{L-1}dp=-\sum_{k=1}^{\color{red}{L}}\frac 1k$$

So that finally $$\boxed{\color{blue}{\gamma \int_0^1(-L)p^{L-1}\ln(1-p)dp}=-\gamma A(L-1)=\color{blue}{\gamma \sum_{k=1}^L \frac 1k}} $$