# Tag Archives: Integral

## Generalized nonlinear polylogarithm integral

\begin{align*} \int^1_0 \frac{\mathrm{Li}_p(x)\,\, \mathrm{Li}_q(x)\, }{x}\, dx&= \sum_{n=1}^{p-1}(-1)^{n-1}\zeta(p-n+1)\zeta(q+n) \\&-\frac{1}{2}\sum_{n=1}^{{p+q}-2}(-1)^{p-1}\zeta(n+1)\zeta({p+q}-n)\\ &+(-1)^{p-1}\left(1+\frac{{p+q}}{2} \right)\zeta({p+q}+1)\end{align*} $$\textit{proof}$$ We can see that $$\int^1_0 \frac{\mathrm{Li}_p(x)\,\, \mathrm{Li}_q(x)\, }{x}\, dx = \sum_{k=1}^\infty\sum_{n=1}^\infty\frac{1}{k^{q}n^{p}(n+k)}$$ Let us first look at the following $$\mathscr{C}(\alpha , k) =\sum_{n=1}^\infty\frac{1}{n^{\alpha}(n+k)}\,\,\, ; \,\,\,\,\mathscr{C}(1, k)=\frac{H_k}{k}$$ This can be … Continue reading

## Integral representation of the zeta function proof

$$\zeta(s) = \frac{1}{\Gamma(s)}\int^\infty_0\frac{t^{s-1}}{e^t-1}dt$$ $$\textit{proof}$$ Start by the integral representation $$\int^\infty_0 \frac{e^{-t}t^{s-1}}{1-e^{-t}}\,dt$$ Using the power expansion $$\frac{1}{1-e^{-t}} = \sum_{n=0}^\infty e^{-nt}$$ Hence we have $$\int^\infty_0\,e^{-t}t^{s-1}\left(\sum_{n=0}^\infty e^{-nt}\right)\,dt$$ By swapping the series and integral $$\sum_{n=0}^\infty\int^\infty_0\,t^{s-1}e^{-(n+1)t}\,dt = \Gamma(s) \sum_{n=0}^\infty \frac{1}{(n+1)^s}=\Gamma(s)\zeta(s)\,$$

## The most ugly looking integral

Prove the following $$I= \log \left\{\frac{\Gamma(b+c+1) \Gamma(c+a+1)\Gamma(a+b+1)}{\Gamma(a+1) \Gamma(b+1) \Gamma(c+1) \Gamma(a+b+c+1)} \right\}$$   where   $$I = \int_0^1 \frac{(1-x^a)(1-x^b)(1-x^c)}{(1-x)(-\log x)}dx$$ $$\textit{proof}$$ First note that since there is a log in the denominator that gives as an idea to use differentiation under … Continue reading

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## Digamma fourth integral representation proof

$$\psi(z) = \log(z) -\frac{1}{2z}-2\int^\infty_0 \frac{t}{(t^2+z^2)(e^{2\pi}-1)}dt\,\,\,\,;\text{ Re}z>0$$ We prove that $$2\int^\infty_0 \frac{t}{(t^2+z^2)(e^{2\pi}-1)}dt= \log(z) -\frac{1}{2z}- \psi(z)$$ First note that $$\frac{2}{e^{2\pi t}-1} =\coth(\pi t)-1$$ Also note that $$\coth(\pi t) = \frac{1}{\pi t}+\frac{2t}{\pi}\sum_{k=1}^\infty\frac{1}{k^2+t^2}$$ Hence we conclude that $$\frac{2t}{e^{2\pi t}-1} =\frac{1}{\pi}-t+\frac{2t^2}{\pi}\sum_{k=1}^\infty\frac{1}{k^2+t^2}$$ Substitute the … Continue reading

## Proof of beta function using convolution

Prove the following $$\beta(x, y)=\int^{1}_{0}t^{x-1}\, (1-t)^{y-1}\,dt= \frac{\Gamma(x)\Gamma{(y)}}{\Gamma{(x+y)}}$$ $$\textit{proof}$$ Let us choose some functions $f$ and $g$ $$f(t) = t^{x} \,\, , \, g(t) = t^y$$ Hence we get $$(t^x*t^y)= \int^{t}_0 s^{x}(t-s)^{y}\,ds$$ So by the convolution rule we have the … Continue reading