Product of polylogarithms and second order polylogarithms

In this post we defined the following

$$\mathscr{H}(p,q) = \sum_{k\geq 1}\frac{\mathscr{C}(p,k)}{k^q}$$

Hence we have the following integral representation

$$\mathscr{H}(p,q)= \int^1_0 \frac{\mathrm{Li}_p(x) \, \mathrm{Li}_q(x)}{x}\, dx$$

\begin{align}
\mathscr{H}(p,q)=\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}

Now for the special case \( p=q\) we have

\begin{align}
\mathscr{H}(q,q)=\int^1_0 \frac{ \mathrm{Li}_q(x) ^2 \, }{x}\, dx&= \sum_{n=1}^{q-1}(-1)^{n-1}\zeta(q-n+1)\zeta(q+n) -\frac{1}{2}\sum_{n=1}^{{2q}-2}(-1)^{q-1}\zeta(n+1)\zeta({2q}-n)\\ &+(-1)^{q-1}\left(1+q \right)\zeta(2q+1)\end{align}

We have for the special case \(q=2\)

$$\mathscr{H}(2,2)=\int^1_0 \frac{ \mathrm{Li}_2(x) ^2 \, }{x}\, dx=2\zeta(3)\zeta(2)-3\zeta(5)$$

Nice Harmonic Identities

Define the following

$$\mathscr{C}(\alpha , k) =\sum_{n\geq 1}\frac{1}{n^{\alpha}(n+k)}\,\,\, ; \,\,\,\,\mathscr{C}(1, k)=\frac{H_k}{k}$$

We can generalize

\begin{align}
\mathscr{C}(\alpha , k) &=\sum_{n\geq 1}\frac{1}{k\, n^{\alpha-1}}\left( \frac{1}{n}-\frac{1}{n+k}\right)\\ &= \frac{1}{k}\zeta(\alpha)-\frac{1}{k}\mathscr{C}(\alpha-1 , k)\\ &=   \frac{1}{k}\zeta(\alpha)-\frac{1}{k^2}\zeta(\alpha-1)+\frac{1}{k^2}\mathscr{C}(\alpha-2 , k)\\ &= \sum_{n=1}^{\alpha-1}(-1)^{n-1}\frac{\zeta(\alpha-n+1)}{k^n}+(-1)^{\alpha-1}\frac{H_k}{k^\alpha}
\end{align}

Hence we have the general formula

$$\mathscr{C}(\alpha , k) = \sum_{n=1}^{\alpha-1}(-1)^{n-1}\frac{\zeta(\alpha-n+1)}{k^n}+(-1)^{\alpha-1}\frac{H_k}{k^\alpha}$$

Dividing by \( \frac{1}{k^{\beta}}\) and summing w.r.t to \(k\) we have

$$\sum_{k\geq 1}\frac{\mathscr{C}(\alpha , k)}{k^{\beta}} = \sum_{n=1}^{\alpha-1}(-1)^{n-1}\zeta(\alpha-n+1)\zeta(\beta+n)+(-1)^{\alpha-1}\sum_{k\geq 1}\frac{H_k}{k^{\alpha+\beta}}$$

Now we use that

$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$

Hence we have

$$\sum_{n=1}^\infty \frac{H_n}{n^{\alpha+\beta}}= \left(1+\frac{{\alpha+\beta}}{2} \right)\zeta({\alpha+\beta}+1)-\frac{1}{2}\sum_{k=1}^{{\alpha+\beta}-2}\zeta(k+1)\zeta({\alpha+\beta}-k)$$

And the generalization is the following formula

\begin{align}
\sum_{k\geq 1}\frac{\mathscr{C}(\alpha , k)}{k^{\beta}} &= \sum_{n=1}^{\alpha-1}(-1)^{n-1}\zeta(\alpha-n+1)\zeta(\beta+n) -\frac{1}{2}\sum_{n=1}^{{\alpha+\beta}-2}(-1)^{\alpha-1}\zeta(n+1)\zeta({\alpha+\beta}-n)\\ &+(-1)^{\alpha-1}\left(1+\frac{{\alpha+\beta}}{2} \right)\zeta({\alpha+\beta}+1)\end{align}

Interestingly we also have that

$$\sum_{k\geq 1}\frac{\mathscr{C}(\alpha , k)}{k^{\beta}}=\sum_{k\geq 1}\frac{\mathscr{C}(\beta, k)}{k^{\alpha}}$$

Generally we will use the symbol

$$\mathscr{H}(\alpha,\beta)=\sum_{k\geq 1}\frac{\mathscr{C}(\alpha , k)}{k^{\beta}}$$

So we have the symmetric property

$$\mathscr{H}(\alpha,\beta) = \mathscr{H}(\beta,\alpha)$$

$$\mathscr{H}(1,\beta)= \left(1+\frac{{\beta}}{2} \right)\zeta({\beta}+1)-\frac{1}{2}\sum_{k=1}^{{\beta}-2}\zeta(k+1)\zeta({\beta}-k)$$

Laplace of an unfamiliar function

Show that 

$$ \int_{0}^{\infty} \, e^{-at} \sin(bt) \frac{\ln t}{t}\, dt =- \left( \frac{ \log(a^2+b^2) }{2} +\gamma \right) \arctan \left( \frac{b}{a} \right)$$

Proof 

We can start by the following integral 

$$I(s) = \int_{0}^{\infty}  t^{s-1} \, e^{-at} \, \sin(bt) dt $$

$$ \int_{0}^{\infty}t^{s-1} \, e^{-at}\sum_{n\geq 0} \frac{(-1)^n (bt)^{2n+1}}{\Gamma(2n+2)} $$

$$  \sum_{n\geq 0} \frac{(-1)^n (b)^{2n+1}}{\Gamma(2n+2)} \int_{0}^{\infty}t^{s+2n} \, e^{-at}  dt$$

$$ \frac{1}{a^s} \sum_{n\geq 0} \frac{(-1)^n (b)^{2n+1} \Gamma (s+2n+1)}{\Gamma(2n+2) a^{2n+1}}$$

By differentiating and plugging \(s=0\) we have 

$$I'(0) = \int_{0}^{\infty} \, e^{-at} \sin(bt) \frac{\ln t}{t}\, dt = \sum_{n\geq 0} \frac{(-1)^n  \psi_0(2n+1)}{2n+1}  \left( \frac{b}{a} \right)^{2n+1} $$$$-\log(a) \sum_{n\geq 0} \frac{(-1)^n }{2n+1}  \left( \frac{b}{a} \right)^{2n+1}$$

$$=  \sum_{n\geq 0} \frac{(-1)^n  H_{2n}-\gamma}{2n+1}  \left( \frac{b}{a} \right)^{2n+1} -\log(a) \arctan \left( \frac{b}{a} \right) $$

$$=  \sum_{n\geq 0} \frac{(-1)^n  H_{2n}}{2n+1}  \left( \frac{b}{a} \right)^{2n+1}-(\gamma +\log(a))\arctan \left( \frac{b}{a} \right) $$

Now we look at the harmonic sum 

$$\sum_{k\geq 0}(-1)^k H_{2k} x^{2k}= \sum_{k\geq 0}(-1)^k x^{2k} \int^1_0 \frac{1-t^{2k}}{1-t} \, dt$$

$$= \int^1_0  \frac{1}{1-t} \sum_{k\geq 0}(-1)^k x^{2k} \left(1-t^{2k}\right) \, dt$$

$$= \int^1_0  \frac{1}{1-t} \sum_{k\geq 0}(-1)^k  \left(x^{2k}-(xt)^{2k}\right) \, dt$$

$$= \int^1_0  \frac{1}{1-t}\left(\frac{1}{1+x^2}-\frac{1}{1+t^2x^2}\right) \, dt$$

$$=\frac{1}{1+x^2} \int^1_0  \frac{1+t^2x^2-1-x^2}{(1-t)(1+t^2x^2)} \, dt$$

$$=\frac{-x^2}{1+x^2} \int^1_0  \frac{(1-t^2)}{(1-t)(1+t^2x^2)} \, dt$$

$$=\frac{-x^2}{1+x^2} \int^1_0  \frac{1+t}{(1+t^2x^2)} \, dt$$

$$=\frac{-x^2}{1+x^2} \left( \int^1_0 \frac{1}{1+t^2x^2}+\frac{t}{1+t^2x^2} \, dt \right) $$

$$=  \frac{-1}{2(1+x^2)} \left(2x \arctan (x) + \log(1+x^2) \right)$$

Using this we conclude by integrating 

$$\sum_{k\geq 0}\frac{(-1)^k H_{2k}}{2k+1} x^{2k}=-\frac{1}{2} \log(1+x^2) \arctan(x)$$

Hence the following 

$$\sum_{k\geq 0}\frac{(-1)^k H_{2k}}{2k+1} \left(\frac{b}{a} \right)^{2k+1}=-\frac{1}{2} \log \left( \frac{a^2+b^2}{a^2} \right) \arctan \left(\frac{b}{a} \right)$$

$$ \int_{0}^{\infty} \, e^{-at} \sin(bt) \frac{\ln t}{t}\, dt  =  -\left( \frac{1}{2} \log \left( \frac{a^2+b^2}{a^2} \right) + \gamma +\log(a) \right) \arctan \left( \frac{b}{a} \right)$$

$$=- \left( \frac{ \log(a^2+b^2) }{2} +\gamma \right) \arctan \left( \frac{b}{a} \right)$$

Contour integration by a sector of a circle

Show that for \( b>1 \) we have

$$ \int^\infty_0 \frac{\ln(x)}{x^b-1}\, dx = \frac{\pi^2}{b^2} \csc^2\left( \frac{\pi}{b}\right)$$

Proof 

Consider the following function

$$f(z) = \frac{\log(z)}{e^{b\log(z)}-1}$$

Now consider a sector with an angle of \(\frac{\pi}{b}\)

The integration of the whole sector is \(0 \) since the function is analytic in and on the contour by taking the principle branch of the logarithm .

The integration along the x-axis is 

$$\int^{R}_{r} \frac{\log(x)}{x^b-1}$$

The integration along the big and small circular sectors 

It can be proven easily by taking \( r\to 0 , R \to \infty\) that they vanish.

The integration along the tilted line

Can be parametrized using the transformation  \(z=te^{i\frac{\pi}{b}}\)  ,  \( R<t<r\)

$$-e^{\frac{\pi i}{b}}\int^{R}_r \frac{\log(te^{\frac{\pi i}{b}})}{e^{b\log\left(\frac{\pi i}{b} \right)}-1}\, dt$$

$$e^{\frac{\pi i}{b}}\int^{R}_r \frac{\log(t)+i\frac{\pi}{b}}{t^b+1}\, dt$$

$$e^{\frac{\pi i}{b}}\int^{R}_r \frac{\log(t)}{t^b+1}\, dt+ie^{\frac{\pi i}{b}}\int^{R}_0\frac{\frac{\pi}{b}}{t^b+1}\, dt\,\,\, (1)$$

The integrals can be easily found using the beta function by taking the principle value and

\(r\to 0 \,\,\,\, , \,\,\,\, R \to \infty\)

$$e^{\frac{\pi i}{b}} \left( -\frac{\pi^2}{b^2} \csc\left( \frac{\pi}{b}\right) \cot\left( \frac{\pi}{b}\right)+i\frac{\pi^2}{b^2} \csc\left( \frac{\pi}{b}\right)\right)$$

This can be written as 

$$-\frac{\pi^2e^{\frac{\pi i}{b}}}{b^2} \csc\left( \frac{\pi}{b}\right) \left(  \cot\left( \frac{\pi}{b}\right)-i\right)$$

$$-\frac{\pi^2e^{\frac{\pi i}{b}}}{b^2} \csc^2\left( \frac{\pi}{b}\right) \left(  \cos\left( \frac{\pi}{b}\right)-i\sin\left( \frac{\pi}{b}\right)\right)$$

Now using the Euler formula we have 

$$-\frac{\pi^2}{b^2} \csc^2\left( \frac{\pi}{b}\right)$$

Summing the curves together and using Cauchy integral formula gives the desired result .

Laplace transform of ugly looking integral

Show that for \( \displaystyle 0 \le a < \frac{\pi}{2}\) ,

$$ \int_{0}^{\infty} e^{-x \cos a} \cos(x \sin a) \cos (bx) \ dx = \frac{(b^{2}+1) \cos a}{b^{4}+2b^{2} \cos (2a) + 1 }$$

Proof 

\begin{align}
\int_{0}^{\infty} e^{-x \cos a} \cos(x \sin a) \cos (bx) \ dx &=  \mathrm{Re}\int_{0}^{\infty} e^{-x \cos a} e^{ix \sin a } \cos (bx) \ dx\\ &= \mathrm{Re}\int_{0}^{\infty} e^{-x e^{-ia}} \cos (bx) \ dx \\& = \mathrm{Re}\frac {e^{-ia}}{e^{-2ia}+b^2}\\&= \mathrm{Re}\frac{\cos(a)-i\sin(a)}{\cos(2a)-i\sin(2a)+b^2}\\&= \mathrm{Re}\frac{(\cos(a)-i\sin(a))(\cos(2a)+b^2+i\sin(2a))}{(\cos(2a)+b^2)^2+\sin^2(2a)}\\&= \frac{\cos(2a)\cos(a)+\cos(a)b^2+\sin(2a)\sin(a)}{b^4+2b^2 \cos(2a)+1}\\&=\frac{(1+b^2)\cos(a)}{b^4+2b^2 \cos(2a)+1}
\end{align}

The convergence is justified by the Laplace transform. Since \( |\cos(bx)| \leq 1\)  so it is of an exponential order and we can take\( |\cos(bx)| \leq e^{0\, x} \) so the value of \(c=0\). Hence the integral converges to the value for \(\mathrm{Re}(e^{ia})>0\) or \(\cos(a)>0\) which clearly satisfy \(0 \leq a < \frac{\pi}{2}\)