×

# Trigamma Series

Evaluate $\sum_{k=1}^{\infty} \left[ \dfrac{\psi^{(1)} (k) }{k} \right]^{2}$

$$\psi^{(1)} (k)$$ is the trigamma function, defined as

$\psi^{(1)} (x) = \dfrac{\mathrm{d}^2}{\mathrm{d}x^2} [\ln (\Gamma (x) )]$

This is a part of the set Formidable Series and Integrals

Note by Ishan Singh
2 years ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Here's a irrelevant comment:::: (Plagiarized Ishu's work)

Claim: $$\displaystyle \sum_{k=1}^\infty ( \psi^{(1)} (k))^2 = 3 \zeta(3)$$.

Proof:

By writing the trigamma function in terms of integral representation, we have $$\displaystyle \psi^{(1)}(k) = \int_0^1 \dfrac{\ln x \cdot x^{k-1}}{1-x} \, dx$$. Taking its square gives us

$\begin{eqnarray} \displaystyle \left [ \psi^{(1)}(k) \right]^2 &= & \displaystyle \int_0^1\int_0^1 \dfrac{ \ln x \ln y (xy)^{k-1}}{(1-x)(1-y) } \, dx dy \\ \displaystyle \sum_{k=1}^\infty \left [ \psi^{(1)}(k) \right]^2 &= & \displaystyle \int_0^1\int_0^1 \dfrac{ \ln x \ln y}{(1-x)(1-y)} \sum_{k=1}^\infty (xy)^{k-1} \, dx dy \\ &= & \displaystyle \int_0^1 \int_0^1 \dfrac { \ln x \ln y}{(1-x)(1-y)(1-xy) } \, dx dy \\ &= & \displaystyle \int_0^1 \dfrac{ \ln x}{1-x} \int_0^1 \dfrac{ \ln y }{(1-y)(1-xy) } \, dy dx \end{eqnarray}$

Consider the integral $\int_0^1 \dfrac{ \ln y }{(1-y)(1-xy) } \, dy = \displaystyle \dfrac y{y-1} \underbrace{\int_0^1 \dfrac {\ln y}{1-xy} \, dy}_{= \, A} - \dfrac1{y-1} \underbrace{\int_0^1 \dfrac{ \ln y}{y - 1} \, dy}_{= \, B}$

Solving for $$A$$ gives

$\begin{eqnarray} \displaystyle \int_0^1 \dfrac {\ln y}{1-xy} \, dy &=& \displaystyle \int_0^1 \ln y \sum_{x=1}^\infty (xy)^{n-1} \, dy \\ &=& \displaystyle \sum_{n=1}^\infty x^{n-1} \int_0^1 \ln y \cdot y^{n-1} \, dy \\ &=& - \displaystyle \sum_{n=1}^\infty \dfrac{x^{n-1}}{n^2} = -\dfrac{ \text{Li}_2 (x)} x \end{eqnarray}$

Similarly, solving for $$B$$ gives $$- \dfrac{\text{Li}_2 (1) }1 = - \dfrac{\pi^2}6$$.

Thus we have $$\displaystyle \int_0^1 \dfrac{ \ln y}{(1-y)(1-xy) } \, dy = \dfrac1{1-x} \left [ \text{Li}_2 (x) - \dfrac{\pi^2} 6 \right ]$$, and so the double integral becomes

$\int_0^1 \dfrac {\ln x}{(1-x)^2 } \left [ \text{Li}_2 (x) - \dfrac{\pi^2} 6 \right ] \, dx = \int_0^1 \dfrac {\ln (1-x)}{x^2 } \left [ \text{Li}_2 (1-x) - \dfrac{\pi^2} 6 \right ] \, dx$

Because $$\displaystyle \int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$$. We integrate by parts with $$du = \dfrac{ \ln(1-x)}{x^2}, v = \text{Li}_2 (x) - \dfrac{\pi^2} 6$$ to get

$u = \int_0^x \dfrac{\ln (1-t)}{t^2} \, dt = -\dfrac{(1-x)}x - \int_0^x \dfrac {dt}{t(1-t)} = \ln(1-x) - \ln(x) - \dfrac{ \ln(1-x)}x$

And $$\dfrac{dv}{dx} = \left [ \text{Li}_2 (x) - \dfrac{\pi^2} 6 \right ] = \dfrac{\ln x}{1-x}$$.

Integrate by parts gives us

$\begin{eqnarray} \displaystyle \int_0^1 \dfrac {\ln (1-x)}{x^2 } \left [ \text{Li}_2 (1-x) - \dfrac{\pi^2} 6 \right ] \, dx &=& - \displaystyle \int_0^1 \dfrac{ \ln x}{1-x} \left [ \ln(1-x) - \ln x - \dfrac{ \ln(1-x)}x \right ] \, dx \\ &=& -\displaystyle \underbrace{ \int_0^1 \dfrac{ \ln x \ln(1-x)}{1-x} \, dx}_{= \, \rm{I}} + \underbrace{ \int_0^1 \dfrac{ (\ln x)^2}{1-x} \, dx}_{= \, \rm{II}}+ \underbrace{ \int_0^1 \dfrac{ \ln x \ln(1-x)}{x(1-x)} \, dx}_{= \, \rm{III}} \end{eqnarray}$

Now we just need to find the values of $$\rm{I}, \rm{II}$$ and $$\rm{III}$$. Taking note that $$B(a,b)$$ is the beta function, we have

$\begin{eqnarray} \displaystyle \rm{I} &=& \displaystyle\int_0^1 \dfrac{ \ln x \ln(1-x)}{1-x} \, dx \\ \displaystyle &=& \displaystyle \lim_{a\to0^+}\lim_{b\to1} \dfrac {\delta}{\delta a} \left( \dfrac {\delta}{\delta b} B(a,b) \right) = \zeta(3) \\ \displaystyle \rm{II} &=& \displaystyle\int_0^1 \dfrac{ (\ln x)^2}{1-x} \, dx \\ \displaystyle &=& \displaystyle \lim_{a\to0^+}\lim_{b\to1} \dfrac {\delta^2}{\delta a^2} \left ( B(a,b) \right) = 2\zeta(3) \\ \displaystyle \rm{III} &=& \displaystyle\int_0^1 \dfrac{ \ln x \ln(1-x)}{x(1-x)} \, dx \\ \displaystyle &=& \displaystyle \lim_{a\to0^+}\lim_{b\to0^+} \dfrac {\delta}{\delta a} \left( \dfrac {\delta}{\delta b} B(a,b) \right) = 2\zeta(3) \end{eqnarray}$

Combining them all together:

$\displaystyle - \int_0^1 \dfrac {\ln (1-x)}{x^2 } \left [ \text{Li}_2 (1-x) - \dfrac{\pi^2} 6 \right ] \, dx =- \zeta(3) + 2\zeta(3) + 2\zeta(3) = \boxed{3\zeta(3)} .$

- 2 years ago

Nice one! It doesn't matter if it is plagiarised or not.

- 2 years ago

Try solving the actual question! =D

- 2 years ago

Comment deleted Mar 13, 2016

Comment deleted Feb 20, 2016

$$\Huge \pi ^{THE CRIMINAL}$$

:P

- 2 years ago

Hey, join here. We talk about integrals and math and integrals.

- 2 years ago

Comment deleted Feb 20, 2016

Comment deleted Feb 20, 2016

Comment deleted Feb 20, 2016

Comment deleted Feb 20, 2016

Comment deleted Mar 13, 2016

Haha ,no worries. Good luck on your test!

- 2 years ago

Comment deleted Feb 20, 2016

Comment deleted Mar 13, 2016

$$\displaystyle \int \int xy dxdy = (x^{2} y^{2})/4$$

Is this right?

- 2 years ago

Comment deleted Mar 13, 2016

KK thanks a lot!

- 2 years ago

$$\displaystyle \frac{93}{24}\zeta(6)$$

- 1 year, 10 months ago

That's incorrect.

- 8 months, 2 weeks ago

Hi! Looks like a calculation error (oops! the sum should also contain some $\zeta(3)^2$ term). Anyway, it was solved here http://integralsandseries.prophpbb.com/topic681.html

- 8 months, 2 weeks ago

Nice. I have discovered another solution which allows us to generalize the result to higher powers and other polygamma functions as well.

- 8 months, 2 weeks ago

×