## Euler Hypergeometric transformation proof

$${}_2F_1 \left(a,b;c;z\right)=(1-z)^{c-a-b} {}_2F_1 \left(c-a,c-b;c;z\right)$$

$$\textit{proof}$$

In the Pfaff transformations let $z \to \frac{z}{z-1}$ , proved here

$${}_2F_1 \left(a,b;c;\frac{z}{z-1}\right)=(1-z)^{-a} {}_2F_1 \left(a,c-b;c;z\right)$$

and

$${}_2F_1 \left(a,b;c;\frac{z}{z-1}\right)=(1-z)^{-b} {}_2F_1 \left(c-a,b;c;z\right)$$

By equating the two transformations

$$(1-z)^{-a} {}_2F_1 \left(a,c-b;c;z\right)=(1-z)^{-b} {}_2F_1 \left(c-a,b;c;z\right)$$

Now use the transformation $b \to c-b$

$$(1-z)^{-a} {}_2F_1 \left(a,b;c;z\right)=(1-z)^{c-b} {}_2F_1 \left(c-a,c-b;c;z\right)$$

Which can be reduced to

$${}_2F_1 \left(a,b;c;z\right)=(1-z)^{c-a-b} {}_2F_1 \left(c-a,c-b;c;z\right)$$

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## Pfaff transformations of the hypergeometric function proof

$${}_2F_1 \left(a,b;c;z\right)=(1-z)^{-a} {}_2F_1 \left(a,c-b;c;\frac{z}{z-1}\right)$$

and

$${}_2F_1 \left(a,b;c;z\right)=(1-z)^{-b} {}_2F_1 \left(c-a,b;c;\frac{z}{z-1}\right)$$

$$\textit{proof}$$

Start by the integral representation

$${}_2F_1 \left(a,b;c;z\right)=\frac{1}{\beta(b,c-b)}\int_0^1 \frac{t^{b-1}(1-t)^{c-b-1}}{(1-tz)^a}\, dt$$

By the the transformation $t\to 1-t$

$$\int_0^1 \frac{(1-t)^{b-1}t^{c-b-1}}{(1-(1-t)z)^a}\, dt=\int_0^1 \frac{(1-t)^{b-1}t^{c-b-1}}{(1-z+tz)^a}\, dt$$

Which can be written as

$$\frac{(1-z)^{-a}}{\beta(b,c-b)}\int_0^1t^{c-b-1}(1-t)^{b-1} \left( 1-t\frac{z}{z-1} \right)^{-a} dt$$

Note this is the integral representation of

$$(1-z)^{-a} {}_2F_1 \left(a,c-b;c;\frac{z}{z-1}\right)\,\,$$

Also using that

$${}_2F_1 \left(a,b;c;z\right)= {}_2F_1 \left(b,a;c;z\right)$$

We deduce that

$${}_2F_1 \left(a,b;c;z\right)=(1-z)^{-b} {}_2F_1 \left(c-a,b;c;\frac{z}{z-1}\right)$$

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## Integral representation of the digamma function using Abel–Plana formula

$$\int^\infty_0 \frac{2x}{(x^2+z^2)e^{2\pi x}-1}\,dx =\log(z)-\psi(z)-\frac{1}{2z}$$

$$\textit{proof}$$

Use Abel–Plana formula

$$\sum_{n=0}^\infty f(n) = \int^\infty_0f(x)\,dx+\frac{f(0)}{2} +i\int^\infty_0 \frac{f(ix)-f(-ix)}{e^{2\pi x}-1}\,dx$$

Let

$$f(x) = \frac{1}{z+x}$$

Note that
$$i(f(ix) -f(-ix))= \frac{i}{z+ix}-\frac{i}{z-ix} = \frac{2x}{z^2+x^2}$$

By integration we have

$$\int^\infty_0 \frac{2x}{(x^2+z^2)e^{2\pi x}-1}\,dx =\lim_{N\to \infty}\sum_{n=0}^N \frac{1}{z+n}-\int^N_0 \frac{1}{x+z}\,dx-\frac{1}{2z}$$

The sum

$$\sum_{n=0}^N \frac{1}{z+n}=\frac{1}{z}+ \sum_{n=1}^N\frac{1}{n}-\frac{z}{n(z+n)} = H_N-\psi(z)-\gamma$$

By taking the limit

\begin{align}
\lim_{N\to \infty}\sum_{n=0}^N \frac{1}{z+n}-\int^N_0 \frac{1}{x+z}\,dx &=\log(z) -\psi(z)-\gamma+\lim_{N\to \infty}H_N-\log(z+N)\\ & = \log(z)-\psi(z)
\end{align}

Finally we have

$$\int^\infty_0 \frac{2x}{(x^2+z^2)e^{2\pi x}-1}\,dx =\log(z)-\psi(z)-\frac{1}{2z}$$

Posted in Digamma function | | 3 Comments

## Sum of natural numbers equal -1/12 ?

$$\tag{1}1+2+3+4+\cdots =^?-\frac{1}{12}$$

There are many problems with that equality. Can a summation of positive integers lead to a negative quantity ?

The equality implies that

$$\sum_{k=1}^\infty k = -\frac{1}{12}$$

But we already know that

$$S_N = \sum_{k=1}^Nk = \frac{N(N+1)}{2} \to {\infty}$$

Clearly the summation is divergent. By all means of divergence tests. More precisely we have $S_N = \mathcal{O}(N^2)$. So the summation acts like a parabola that extends to infinity as we increase the number of terms in the summation.

So, what is the origin of (1) ?

Let us look at the zeta function defined as

$$\zeta(s) = \sum_{k=1}^\infty \frac{1}{k^s}$$

We see that

$$\zeta(-1) = \sum_{k=1}^\infty \frac{1}{k^{-1}} = \sum_{k=1}^\infty k$$

Now according to the functional equation

$$\zeta(s)=2^s\pi^{s-1}\sin \left( \frac{s\pi}2\right) \Gamma(1-s)\zeta(1-s)$$

Then clearly

$$\zeta(-1)=-\frac{1}{2\pi^2}\Gamma(2)\zeta(2)$$

It is well know that $\zeta(2) = \frac{\pi^2}{6}$ and $\Gamma(2) = 1! = 1$

$$\zeta(-1)=-\frac{1}{2\pi^2}\Gamma(2)\zeta(2) = -\frac{1}{12}$$

So we have

$$1+2+3\cdots = -\frac{1}{12}$$

What is wrong with this proof ?!

The problem is that $\zeta(s)$ is differentiable everywhere except at $s = 1$. But the summation formula is not actually convergent unless for $s >1$

Hence

$$\zeta(s) = \sum_{k=1}^\infty \frac{1}{n^s}\,\,;\,\, s > 1$$

It doesn’t make sense to say that

$$\zeta(-1) =^? \sum_{k=1}^\infty k$$

This is because the series definition is not defined for all values just for some values but the functional equation holds for $s\neq 1$.

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## Integral representation of Gauss Hypergeometric function proof

$$\beta(c-b,b) \, _2F_1(a,b;c;z)=\int_0^1 \frac{t^{b-1}(1-t)^{c-b-1}}{(1-tz)^a}\, dt$$

$$\textit{proof}$$

Start by the RHS

$$\int_0^1 t^{b-1}(1-t)^{c-b-1} \, (1-tz)^{-a}\, dt$$

Using the expansion of $(1-tz)^{-a}$ we have

$$\int_0^1 t^{b-1}(1-t)^{c-b-1} \sum_{k=0}^\infty\frac{(a)_k}{k!}\, (tz)^k$$

Interchanging the integral with the series

$$\sum_{k=0}^\infty\frac{(a)_k}{k!}\, z^k \, \int_0^1 t^{k+b-1}(1-t)^{c-b-1}\, dt$$

Recalling the beta function we have

$$\sum_{n=0}^\infty\frac{(a)_k \Gamma(k+b) \Gamma(c-b)}{ \Gamma(k+c)}\, \frac{z^k}{k!}$$

Using the identity that

$$\beta(c-b,b) = \frac{\Gamma(b)\Gamma(c-b)}{\Gamma(c)}$$

and

$$\frac{\Gamma(z+k)}{\Gamma(z)} = (z)_k$$

We deduce that

$$\beta(c-b,b)\sum_{k=0}^\infty\frac{(a)_k \Gamma(k+b) \Gamma(c)}{\Gamma(b) \Gamma(k+c)}\, \frac{z^k}{k!}= \beta(c-b, b) \sum_{k=0}^\infty \frac{(a)_k (b)_k}{(c)_k} \frac{z^k}{k!} \,\,$$

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## Symmetry formula for Generalized Linear Euler sums

$$\sum_{k=1}^\infty \frac{H^{(p)}_k}{k^q}+\sum_{k=1}^\infty \frac{H^{(q)}_k}{k^p} =\zeta(p)\zeta(q)+\zeta(p+q)$$

$$\textit{proof}$$
Take the leftmost series and swap the finite and infinite sums

$$\sum_{i=1}^\infty \,\sum_{k=i}^\infty\frac{1}{i^p} \frac{1}{k^q}=\sum_{i=1}^\infty \,\sum_{k=1}^\infty\frac{1}{i^p} \frac{1}{k^q}-\sum_{i=1}^\infty\frac{1}{i^p} \,\sum_{k=1}^{i-1} \frac{1}{k^q}$$

The second sum can be written as

$$\sum_{i=1}^\infty\frac{1}{i^p} \,\sum_{k=1}^{i-1} \frac{1}{k^q} = \sum_{i=1}^\infty\frac{1}{i^p} \,\left(\sum_{k=1}^{i} \frac{1}{k^q}-\frac{1}{i^p}\right)$$

By separating and changing the index we get

$$\sum_{k=1}^\infty\frac{H^{(q)}_k}{k^p}-\zeta(p+q)$$

Hence we have

$$\sum_{k=1}^\infty \frac{H^{(p)}_k}{k^q} =\zeta(p)\zeta(q)-\sum_{k=1}^\infty\frac{H^{(q)}_k}{k^p}+\zeta(p+q)$$

Which reduces to
$$\sum_{k=1}^\infty \frac{H^{(p)}_k}{k^q}+\sum_{k=1}^\infty\frac{H^{(q)}_k}{k^p} =\zeta(p)\zeta(q)+\zeta(p+q)$$

For the special case $p=q=n$

$$\sum_{k=1}^\infty \frac{H^{(n)}_k}{k^n} =\frac{\zeta^2(n)+\zeta(2n)}{2}$$

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## Integral representation of generalized Euler sums

$$\sum_{k=1}^\infty\frac{H^{(p)}_k}{k^q} = \zeta(p)\zeta(q) +(-1)^{p}\frac{1}{ (p-1)!}\int^1_0\frac{\mathrm{Li}_q(x)\log(x)^{p-1}}{1-x}\,dx$$

$$\textit{proof}$$

Note that

$$\psi_0(a+1)= \int^1_0\frac{1-x^a}{1-x}\,dx$$

By differentiating with respect to $a$ , $p$ times we have

$$\psi_p(a+1) = \frac{\partial}{\partial a^p}\int^1_0\frac{1-x^a}{1-x}\,dx$$

$$\psi_p(a+1) = -\int^1_0\frac{x^a\log(x)^{p}}{1-x}\,dx$$

Let $a =k$

$$\psi_{p-1}(k+1) = -\int^1_0\frac{x^k\log(x)^{p-1}}{1-x}\,dx$$

Use the relation to polygamma

$$H^{(p)}_k = \zeta(p) +(-1)^{p-1}\frac{\psi_{p-1}(k+1)}{ (p-1)!}$$

Hence we have

$$H^{(p)}_k = \zeta(p) +(-1)^{p}\frac{1}{ (p-1)!}\int^1_0\frac{x^k\log(x)^{p-1}}{1-x}\,dx$$

Now divide by $k^q$ and sum with respect to $k$

$$\sum_{k=1}^\infty\frac{H^{(p)}_k}{k^q} = \zeta(p)\zeta(q) +(-1)^{p}\frac{1}{ (p-1)!}\int^1_0\frac{\mathrm{Li}_q(x)\log(x)^{p-1}}{1-x}\,dx$$

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## 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 solved using

\begin{align*}
\mathscr{C}(\alpha , k) &=\sum_{n=1}^\infty\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)\\
&\,\,\cdot \\
&\,\,\cdot \\
&\,\,\cdot \\
&=\frac{1}{k}\zeta(\alpha)-\frac{1}{k^2}\zeta(\alpha-1)+\cdots +(-1)^{\alpha} \frac{\zeta(2)}{k^{\alpha-1}}+\frac{1}{k^{\alpha-1}}\mathscr{C}(\alpha-(\alpha-1) , 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 $k^{\beta}$ and summing w.r.t to $k$

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

Now we use the general formula

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

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=1}^\infty\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*}

We conclude by putting that

$$\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{1}{k^{q}n^{p}(n+k)} = \sum_{k=1}^\infty\frac{\mathscr{C}(p, k)}{k^{q}}$$

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## Nonlinear Euler sums using Nielsen formula

According to Nielsen we have the following :

If $$f(x)= \sum_{n= 0}^\infty a_n x^n$$

Then we have the following

$$\tag{1}\int^1_0 f(xt)\, \mathrm{Li}_2(t)\, dx=\frac{\pi^2}{6x}\int^x_0 f(t)\, dt -\frac{1}{x}\sum_{n=1}^\infty \frac{a_{n-1} H_{n}}{n^2}x^n$$

Now let $a_n = H_n$ then we have the following

$$f(x)=\sum_{n=1}^\infty H_n x^n=-\frac{\log(1-x)}{1-x}$$

$$-\int^1_0 \frac{\log(1-xt)}{1-xt} \mathrm{Li}_2(t)\, dx=-\frac{\pi^2}{6x}\int^x_0 \frac{\log(1-t)}{1-t} dt-\sum_{n=1}^\infty \frac{H_{n-1} H_{n}}{n^2}x^{n-1}$$

Hence we have the following by gathering the integrals and $x\to 1$

$$\sum_{n=1}^\infty \frac{H_{n-1} H_{n}}{n^2}=\int^1_0\frac{\log(1-x)\left(\mathrm{Li}_2(x)-\zeta(2)\right)}{1-x} dx$$

Integrating by parts we have

$$\sum_{n=1}^ \infty \frac{H_{n-1} H_{n}}{n^2}=-\frac{1}{2}\int^1_0\frac{\log(1-x)^3}{x} dx$$

Hence we have

$$\sum_{n=1}^\infty\frac{ H^2_{n}}{n^2}=\sum_{n=1}^\infty \frac{ H_{n}}{n^3}-\frac{1}{2}\int^1_0\frac{\log(1-x)^3}{x} dx=\frac{17 \pi^4}{360}$$

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## Nonlinear euler sum proof using stirling numbers of the first kind

Prove that

$$\sum_{k=1}^\infty \frac{(H_k)^2}{k^2} = \frac{17\pi^4}{360}$$

$$\textit{proof}$$

Start by the following which can be proved by induction

$$\frac{\left[n\atop 3\right]}{n!} =\frac{ (H_{n-1})^2-H^{(2)}_{n-1}}{2n}$$

And the generating function proved here

$$-\sum_{n=3}^\infty \left[n\atop 3\right] \frac{z^n}{n!} = \frac{\log^3(1-z)}{6}$$

Hence we get

$$\sum_{n=3}^\infty ( H^{(2)}_{n-1}- (H_{n-1})^2) \frac{z^n}{n}= \frac{\log^3(1-z)}{3}$$

Which implies that

$$\sum_{n=3}^\infty \frac{H^{(2)}_{n-1}- (H_{n-1})^2}{n^2}= \frac{1}{3}\int^1_0\frac{\log^3(1-z)}{z}\,dz$$

By integration

$$\sum_{n=3}^\infty \frac{(H_{n-1})^2}{n^2}=\sum_{n=3}^\infty \frac{H^{(2)}_{n-1}}{n^2}-\frac{1}{3}\int^1_0\frac{\log^3(1-z)}{z}\,dz$$

Rewritten as

$$\sum_{n=3}^\infty \frac{(H_{n})^2}{n^2}-2\sum_{n=3}^\infty \frac{H_n}{n^3}+\sum_{n=3}^\infty \frac{1}{n^4}=\sum_{n=3}^\infty \frac{H^{(2)}_{n-1}}{n^2}-\frac{1}{3}\int^1_0\frac{\log^3(1-z)}{z}\,dz$$

Finally we have

$$\sum_{n=3}^\infty \frac{(H_{n})^2}{n^2}=2\sum_{n=3}^\infty \frac{H_n}{n^3}-2\sum_{n=3}^\infty \frac{1}{n^4}+\sum_{n=3}^\infty \frac{H^{(2)}_{n}}{n^2}-\frac{1}{3}\int^1_0\frac{\log^3(1-z)}{z}\,dz$$

All sums are shifted known terms

$$\sum_{n=1}^\infty \frac{(H_{n})^2}{n^2}=\left(H^{2}_{1}+\frac{H^{2}_{2}}{4} \right)-2\left(H_1+\frac{H_2}{8} \right)+2\left(1+\frac{1}{16}\right)-\left(H^{(2)}_{1}+\frac{H^{(2)}_{2}}{4} \right) \\+2\sum_{n=1}^\infty \frac{H_n}{n^3}-2\zeta(4)+\sum_{n=1}^\infty \frac{H^{(2)}_{n}}{n^2}-\frac{1}{3}\int^1_0\frac{\log^3(1-z)}{z}\,dz$$

This simplifies to

$$\sum_{n=1}^\infty \frac{(H_{n})^2}{n^2}=2\sum_{n=1}^\infty \frac{H_n}{n^3}-2\zeta(4)+\sum_{n=1}^\infty \frac{H^{(2)}_{n}}{n^2}-\frac{1}{3}\int^1_0\frac{\log^3(1-z)}{z}\,dz$$
For the Euler sums we use

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

And

$$\sum_{n= 1}^\infty \frac{H^{(k)}_n}{n^k}\, = \frac{\zeta{(2k)}+\zeta^{2}(k)}{2}$$

Then we have

$$\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^2} = \frac{7\pi^4}{360}$$

And

$$\sum_{n=1}^\infty \frac{H_n^{(1)}}{n^3} = \frac{\pi^4}{72}$$

Finally we prove that

$$\int^1_0 \frac{\log^3(1-x)}{x}\,dx = -6\zeta(4)$$

Note that

$$\int^1_0 \frac{\log^3(1-x)}{x}\,dx = \sum_{n=0}^\infty \int^1_0 x^{n}\log^3(x)\,dx =-6\sum_{n=0}^\infty \frac{1}{(n+1)^4} = -6\zeta(4)$$

Collecting all the results we get our result

$$\sum_{n=1}^\infty \frac{(H_{n})^2}{n^2}=\frac{2\pi^4}{72}-2\zeta(4)+ \frac{7\pi^4}{360}+2\zeta(4) = \frac{17\pi^4}{360}$$

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