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

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

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:

\[\displaystyle - \int_0^1 \dfrac {\ln (1-x)}{x^2 } \left [ \text{Li}_2 (1-x) - \dfrac{\pi^2} 6 \right ] \, dx =- \zeta(3) + 2\zeta(3) + 2\zeta(3) = \boxed{3\zeta(3)} . \] – Pi Han Goh · 1 year, 7 months ago

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– Aditya Kumar · 1 year, 7 months ago

Nice one! It doesn't matter if it is plagiarised or not.Log in to reply

– Pi Han Goh · 1 year, 7 months ago

Try solving the actual question! =DLog in to reply

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:P– Harsh Shrivastava · 1 year, 7 months agoLog in to reply

join here. We talk about integrals and math and integrals. – Pi Han Goh · 1 year, 7 months ago

Hey,Log in to reply

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– Pi Han Goh · 1 year, 7 months ago

Haha ,no worries. Good luck on your test!Log in to reply

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Is this right? – Harsh Shrivastava · 1 year, 7 months ago

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– Harsh Shrivastava · 1 year, 7 months ago

KK thanks a lot!Log in to reply

\(\displaystyle \frac{93}{24}\zeta(6)\) – Ramya Datta · 1 year, 5 months ago

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– Ishan Singh · 3 months, 2 weeks ago

That's incorrect.Log in to reply

– Ramya Datta · 3 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.htmlLog in to reply

– Ishan Singh · 3 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.Log in to reply