A FAMILY OF LOG INTEGRALS - 2

Today´s post I want to proof the following integrals.



\begin{aligned} &\int_0^1\frac{x^{z-1}-1}{\ln x}dx=\ln z\\ & \\ &\int_0^1\left(\frac{x^{z-1}}{\ln x}+\frac{x^{w-1}}{1-x} \right) dx=\ln z-\psi(w)\\ &\\ &\int_0^{1}\frac{x^{z-1}-x^{w-1}}{1-x}dx=\psi(w)-\psi(z)\\ & \\ & \int_0^{\infty}\frac{x^{z-1}-x^{w-1}}{1-x}dx=\pi \cot(\pi z)-\pi \cot(\pi w)\\ & \\ &{\int_0^{\infty}\frac{x^{z-1}\ln x}{1-x^n}dx=-\frac{\pi^2}{n^2 }\csc^2 \left( \frac{\pi z}{n} \right)\\ &\\ &\int_0^{\infty}\frac{x^{z-1}\ln^2 x}{1-x^n}dx=\frac{2\pi^3}{n^3}csc^{2}\left( \frac{\pi z}{n}\right)\cot\left( \frac{\pi z}{n} \right)\\ & \\ &\int_0^{\infty}\frac{x^{z-1}\ln^3 x}{1-x^n}dx=-\frac{2\pi^4}{n^4} \left(\csc^2\left( \frac{\pi z}{n} \right)+ 3\csc^{2}\left( \frac{\pi z}{n} \right)\cot^2\left( \frac{\pi z}{n} \right)\right) \end{aligned}

First, recall the Digamma integral representation


\psi(z)=\int_{0}^{\infty} \frac{e^{-t}}{t}-\frac{e^{-z t}}{1-e^{-t}} d t

\text {by the following substitution } e^{-t}=x \Rightarrow d t=-\frac{d x}{x} \,\text {we get }

\psi(z)=\int_{1}^{0} \frac{x}{-\ln x}-\frac{x^{z}}{1-x} \frac{(-d x)}{x}

\boxed{\psi(z)=-\int_{0}^{1} \frac{1}{\ln x}+\frac{x^{z-1}}{1-x} d x}(1)

Let´s now evaluate the following integral

I(w)=\int_0^1\frac{x^{z-1}-1}{\ln x}dx=\ln z

I(w)=\int_0^1\frac{x^{z-1}-1}{\ln x}dx=\int_0^1\frac{e^{(z-1)\ln x}-1}{\ln x}dx

I^{\prime}(w)=\int_0^1\frac{x^{z-1} \ln x}{\ln x}dx=\int_0^1x^{z-1}dx

I^{\prime}(w)=\frac{1}{z}

\int_0^1\frac{x^{z-1}-1}{\ln x}dx=\ln z+ C

Setting z=1 we obtain C=0

\boxed{\int_0^1\frac{x^{z-1}-1}{\ln x}dx=\ln z}(2)

\int_0^1\left(\frac{x^{z-1}}{\ln x}+\frac{x^{w-1}}{1-x} \right) dx=\ln z-\psi(w)

\int_0^1\left(\frac{x^{z-1}}{\ln x}+\frac{x^{w-1}}{1-x} \right) dx=\int_0^1\left(\frac{1}{\ln x}-\frac{1}{\ln x}+\frac{x^{z-1}}{\ln x}+\frac{x^{w-1}}{1-x} \right) dx

=\int_0^1\frac{x^{z-1}-1}{\ln x}dx+\int_0^1\left(\frac{1}{\ln x}+\frac{x^{w-1}}{1-x} \right) dx

From (1) and (2) we obtain

\boxed{\int_0^1\left(\frac{x^{z-1}}{\ln x}+\frac{x^{w-1}}{1-x} \right) dx=\ln z-\psi(w)}(3)

I=\int_0^{1}\frac{x^{z-1}-x^{w-1}}{1-x}dx

I=\int_0^{1}\left(\frac{x^{z-1}-x^{w-1}}{1-x}+\frac{1}{\ln x}-\frac{1}{\ln x}\right)dx

I=\int_0^{1}\frac{1}{\ln x}+\frac{x^{w-1}}{1-x}dx-\int_0^{1}\frac{1}{\ln x}+\frac{x^{z-1}}{1-x}dx

From (2) we get

\boxed{\int_0^{1}\frac{x^{z-1}-x^{w-1}}{1-x}dx=\psi(w)-\psi(z)}(4)

I=\int_0^{\infty}\frac{x^{z-1}-x^{w-1}}{1-x}dx

=\int_0^{1}\frac{x^{z-1}-x^{w-1}}{1-x}dx+\int_1^{\infty}\frac{x^{z-1}-x^{w-1}}{1-x}dx

Let x=\frac{1}{t} in the second integral

=\int_0^{1}\frac{x^{z-1}-x^{w-1}}{1-x}dx+\int_1^{0}\frac{t^{1-z}-t^{1-w}}{1-\frac{1}{t}}\frac{(-dt)}{t^2}

=\int_0^{1}\frac{x^{z-1}-x^{w-1}}{1-x}dx+\int_0^1\frac{t^{1-z}-t^{1-w}}{t(t-1)}dt

=\int_0^{1}\frac{x^{z-1}-x^{w-1}}{1-x}dx-\int_0^1\frac{t^{-z}-t^{-w}}{1-t}dt

=\left(\psi(w)-\psi(z)\right)-\left(\psi(1-w)-\psi(1-z)\right)

=\left(\psi(w)-\psi(1-w)\right)-\left(\psi(z)\right)-\psi(1-z)\right)

\boxed{\int_0^{\infty}\frac{x^{z-1}-x^{w-1}}{1-x}dx=\pi \cot(\pi z)-\pi \cot(\pi w)}(5)

Consider (5)

I\left(z,w \right)=\int_0^{\infty}\frac{x^{z-1}-x^{w-1}}{1-x}dx

I\left(z,w \right)=\int_0^{\infty}\frac{e^{(z-1)\ln x}-e^{(w-1)\ln x}}{1-x}dx

\frac{\partial I\left(z,w \right)}{\partial z}=\int_0^{\infty}\frac{e^{(z-1)\ln x}-e^{(w-1)\ln x}}{1-x}dx

\frac{\partial I\left(z,w \right)}{\partial z}=\int_0^{\infty}\frac{x^{z-1}\ln x}{1-x}dx(6)

On the other hand

I\left(z,w \right)=\pi \cot(\pi z)-\pi \cot(\pi w)}

\frac{\partial I\left(z,w \right)}{\partial z}=-\pi^2 \csc^2 \left( \pi z \right)(7)

Equating (6) and (7)

\boxed{\int_0^{\infty}\frac{x^{z-1}\ln x}{1-x}dx=-\pi^2 \csc^2 \left( \pi z \right)}(8)

Consider now a more general version of (8)

I=\int_0^{\infty}\frac{x^{z-1}\ln x}{1-x^n}dx

Let x^n=t, then

I=\frac{1}{n}\int_0^{\infty}\frac{t^{\frac{z-1}{n}}\ln t^{\frac{1}{n}}}{1-t}t^{\frac{1}{n}-1}dt

I=\frac{1}{n^2 }\int_0^{\infty}\frac{t^{\frac{z}{n}-1}\ln t}{1-t}dt

Therefore, form (8) we obtain

\boxed{\int_0^{\infty}\frac{x^{z-1}\ln x}{1-x^n}dx=-\frac{\pi^2}{n^2 }\csc^2 \left( \frac{\pi z}{n} \right)}(9)

I(z)=\int_0^{\infty}\frac{x^{z-1}\ln x}{1-x}dx

I^{\prime}(z)=\int_0^{\infty}\frac{x^{z-1}\ln^2 x}{1-x}dx(10)

On the other hand

I(z)=-\pi^2 \csc^2\left( \pi z \right)

I(z)=-\pi^2 \sin^{-2}\left( \pi z \right)

I^{\prime}(z)=2\pi^3 \sin^{-3}\left( \pi z \right)\cos\left( \pi z \right)

I^{\prime}(z)=2\pi^3 \csc^{2}\left( \pi z \right)\cot\left( \pi z \right)(11)

Equating (10) and (11)

\boxed{\int_0^{\infty}\frac{x^{z-1}\ln^2 x}{1-x}dx=2\pi^3 \csc^{2}\left( \pi z \right)\cot\left( \pi z \right)}(12)

I(n)=\int_0^{\infty}\frac{x^{z-1}\ln^2 x}{1-x^n}dx

Let x=t^{\frac{1}{n}} then

I(n)=\frac{1}{n^3}\int_0^{\infty}\frac{x^{\frac{z}{n}-1}\ln^2 t}{1-t}dx

and from (12) we obtain

\boxed{\int_0^{\infty}\frac{x^{z-1}\ln^2 x}{1-x^n}dx=\frac{2\pi^3}{n^3}csc^{2}\left( \frac{\pi z}{n}\right)\cot\left( \frac{\pi z}{n} \right)}(13)

I(z)=\int_0^{\infty}\frac{x^{z-1}\ln^2 x}{1-x}dx

I^{\prime}(z)=\int_0^{\infty}\frac{x^{z-1}\ln^3 x}{1-x}dx(14)

I(z)=2\pi^3 \frac{\cos(\pi z)}{\sin^3(\pi z)}

I^{\prime}(z)=2\pi^3 \left(\frac{-\pi\sin^4(\pi z)-3\pi \sin^2(\pi z) \cos^2(\pi z)}{\sin^6(\pi z)}\right)

I^{\prime}(z)=-2\pi^4 \left(\frac{1}{\sin^2(\pi z)}+ 3\frac{\cos^2(\pi z)}{\sin^4(\pi z)}\right)

I^{\prime}(z)=-2\pi^4 \left(\csc^2(\pi z)+ 3\csc^{2}\left( \pi z \right)\cot^2\left( \pi z \right)\right)(15)

Equating (14) and (15)

\boxed{\int_0^{\infty}\frac{x^{z-1}\ln^3 x}{1-x}dx=-2\pi^4 \left(\csc^2(\pi z)+ 3\csc^{2}\left( \pi z \right)\cot^2\left( \pi z \right)\right)}(16)

I(n)=\int_0^{\infty}\frac{x^{z-1}\ln^3 x}{1-x^n}dx

Let x=t^{\frac{1}{n}} then

I(n)=\frac{1}{n^4}\int_0^{\infty}\frac{x^{\frac{z}{n}-1}\ln^3 t}{1-t}dx(17)

\boxed{\int_0^{\infty}\frac{x^{z-1}\ln^3 x}{1-x^n}dx=-\frac{2\pi^4}{n^4} \left(\csc^2\left( \frac{\pi z}{n} \right)+ 3\csc^{2}\left( \frac{\pi z}{n} \right)\cot^2\left( \frac{\pi z}{n} \right)\right)}(18)

Comments

Post a Comment

Popular posts from this blog