# 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

$$\psi \left(a\right)=\int^{\infty}_0 \, \frac{e^{-t}}{t}-\frac{e^{-\left(a t\right)}}{1-e^{-t}}\, dt$$ $$\textit{proof}$$ Let $e^{-t}=x$ $$\int^{1}_0 \, -\frac{1}{\log(x)}-\frac{x^{a-1}}{1-x}\, dx$$ By adding and subtracting 1 $$-\int^{1}_0 \, \frac{1}{\log(x)}+\frac{1}{1-x}\, dx+\int^1_0\frac{1-x^{a-1}}{1-x}\, dx$$ Using the second integral representation $$-\int^{1}_0 \, \frac{1}{\log(x)}+\frac{1}{1-x}\, dx+\gamma+\psi(a)$$ We can prove that $$\int^{1}_0 \, … Continue reading Posted in Digamma function | Tagged , , | Leave a comment ## Second integral representation of digamma proof$$\psi(s+1)\,=\, -\gamma \,+\, \int^{1}_{0}\frac{1-x^s}{1-x}\,dx\textit{proof}$$This can be done by noting that$$\psi(s+1) = -\gamma +\sum_{n=1}^\infty\frac{s}{n(n+s)}$$It is left as an exercise to prove that$$\sum_{n=1}^\infty\frac{s}{n(n+s)} = \int^{1}_{0}\frac{1-x^s}{1-x}\,dx$$Posted in Digamma function | Tagged , , | Leave a comment ## First integral representation of digamma proof$$ \psi(a) = \int^{\infty}_0 \frac{e^{-z}-(1+z)^{-a}}{z}\,dz \textit{proof}$$We begin with the double integral$$\int^{\infty}_0 \int^t_1 \,e^{-xz}\,dx\,dz=\int^{\infty}_{0}\frac{e^{-z}-e^{-tz}}{z}\, dz $$Using fubini theorem we also have$$ \int^t_1 \int^{\infty}_0\,e^{-xz}\,dz \,dx = \int^t_1 \frac{1}{x}\,dx = \log t $$Hence we have the following … Continue reading Posted in Digamma function | Tagged , , | Leave a comment ## Cosine sine integral Prove that$$\int_0^{2\pi} \cos(\sin{x})e^{\cos{x}} dx = 2\pi\textit{proof}\begin{align} \int_0^{2\pi} \cos(\sin{x})e^{\cos{x}} dx &=Re\left( \int_0^{2\pi} e^{i\sin(x)}e^{\cos{x}} dx\right) \\ &= Re \left(\int_0^{2\pi} e^{e^{i x}} dx\right)\\ &= Re \left(\sum_{n=0}^\infty \frac{1}{n!}\int_0^{2\pi} e^{inx} dx\right) \\ &= \sum_{n=0}^\infty \frac{1}{n!}\int_0^{2\pi} \cos(nx) dx \\ &= \int_0^{2\pi} dx = … Continue reading
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