Trigamma Series

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

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

ψ(1)(x)=d2dx2[ln⁡(Γ(x))] \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

#Calculus

Note by Ishan Singh
5 years, 3 months ago

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Comments

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

Claim: ∑k=1∞(ψ(1)(k))2=3ζ(3) \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 ψ(1)(k)=∫01ln⁡x⋅xk−11−x dx \displaystyle \psi^{(1)}(k) = \int_0^1 \dfrac{\ln x \cdot x^{k-1}}{1-x} \, dx . Taking its square gives us

[ψ(1)(k)]2=∫01∫01ln⁡xln⁡y(xy)k−1(1−x)(1−y) dxdy∑k=1∞[ψ(1)(k)]2=∫01∫01ln⁡xln⁡y(1−x)(1−y)∑k=1∞(xy)k−1 dxdy=∫01∫01ln⁡xln⁡y(1−x)(1−y)(1−xy) dxdy=∫01ln⁡x1−x∫01ln⁡y(1−y)(1−xy) dydx \begin{aligned} \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{aligned}

Consider the integral ∫01ln⁡y(1−y)(1−xy) dy=yy−1∫01ln⁡y1−xy dy⏟= A−1y−1∫01ln⁡yy−1 dy⏟= B\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 AA gives

∫01ln⁡y1−xy dy=∫01ln⁡y∑x=1∞(xy)n−1 dy=∑n=1∞xn−1∫01ln⁡y⋅yn−1 dy=−∑n=1∞xn−1n2=−Li2(x)x \begin{aligned} \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{aligned}

Similarly, solving for BB gives −Li2(1)1=−π26 - \dfrac{\text{Li}_2 (1) }1 = - \dfrac{\pi^2}6 .

Thus we have ∫01ln⁡y(1−y)(1−xy) dy=11−x[Li2(x)−π26] \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

∫01ln⁡x(1−x)2[Li2(x)−π26] dx=∫01ln⁡(1−x)x2[Li2(1−x)−π26] dx \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 ∫abf(x) dx=∫abf(a+b−x) dx \displaystyle \int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx . We integrate by parts with du=ln⁡(1−x)x2,v=Li2(x)−π26du = \dfrac{ \ln(1-x)}{x^2}, v = \text{Li}_2 (x) - \dfrac{\pi^2} 6 to get

u=∫0xln⁡(1−t)t2 dt=−(1−x)x−∫0xdtt(1−t)=ln⁡(1−x)−ln⁡(x)−ln⁡(1−x)x 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 dvdx=[Li2(x)−π26]=ln⁡x1−x\dfrac{dv}{dx} = \left [ \text{Li}_2 (x) - \dfrac{\pi^2} 6 \right ] = \dfrac{\ln x}{1-x} .

Integrate by parts gives us

∫01ln⁡(1−x)x2[Li2(1−x)−π26] dx=−∫01ln⁡x1−x[ln⁡(1−x)−ln⁡x−ln⁡(1−x)x] dx=−∫01ln⁡xln⁡(1−x)1−x dx⏟= I+∫01(ln⁡x)21−x dx⏟= II+∫01ln⁡xln⁡(1−x)x(1−x) dx⏟= III \begin{aligned} \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{aligned}

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

I=∫01ln⁡xln⁡(1−x)1−x dx=lim⁡a→0+lim⁡b→1δδa(δδbB(a,b))=ζ(3)II=∫01(ln⁡x)21−x dx=lim⁡a→0+lim⁡b→1δ2δa2(B(a,b))=2ζ(3)III=∫01ln⁡xln⁡(1−x)x(1−x) dx=lim⁡a→0+lim⁡b→0+δδa(δδbB(a,b))=2ζ(3) \begin{aligned} \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{aligned}

Combining them all together:

−∫01ln⁡(1−x)x2[Li2(1−x)−π26] dx=−ζ(3)+2ζ(3)+2ζ(3)=3ζ(3).\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 - 5 years, 3 months ago

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

Aditya Kumar - 5 years, 3 months ago

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Try solving the actual question! =D

Pi Han Goh - 5 years, 3 months ago

9324ζ(6)\displaystyle \frac{93}{24}\zeta(6)

Ramya Datta - 5 years, 2 months ago

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That's incorrect.

Ishan Singh - 4 years ago

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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

Ramya Datta - 4 years ago

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

Ishan Singh - 4 years ago
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