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This is @Amad27 something happened to my account, which I will get fixed soon, so for now I will ask as a guest until the problem is fixed. Thanks.

I saw this method of calculating:

$$I = \int_{0}^{1} \log^2(1-x)\log^2(x) dx$$

Evaluate $\int^1_0 \log^2(1-x) \log^2(x) \, dx$

Can you take a look at M.N.C.E.'s method?

I dont understand a few things.

Somehow he makes the relation:

$$\frac{4H_n}{(n+1)(n+2)^3} = \frac{\left( \gamma + \psi(-z) \right)^2}{(z+1)(z+2)^3}$$

How is this established?

And this I dont understand, why did he integrate it,?

And then after he states: "At the positive integers," what is he doing with the residues. I know the residue theorem etc, but I dont understand what he is exactly doing? Thanks

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amad27
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It is not the case that $$ \frac{4H_n}{(n+1)(n+2)^3}=\frac{(\gamma+\psi(-z))^2}{(z+1)(z+2)^3} $$ What is being used are the residues of $f(z)=\frac{(\gamma+\psi(-z))^2}{(z+1)(z+2)^3}$.

Using the relation $$ \gamma+\psi(-z)=\sum_{k=0}^\infty\left(\frac1{k+1}+\frac1{z-k}\right) $$ we get that the residue of $f(z)$ at $z=n\in\mathbb{Z}$ for $n\ge0$ is $$ \frac{2H_n}{(n+1)(n+2)^3}-\frac{4n+5}{(n+1)^2(n+2)^4} $$ Furthermore, we get that the residue of $f(z)$ at $z=-1$ is $0$ and at $z=-2$ is $$ 2\zeta(3)-2-\left(2-\frac{\pi^2}6\right)^2 $$ On a path that stays away from the negative integers (say more than $\frac13$ away), $|f(z)|\sim\frac{\log(|z|)^2}{|z|^4}$. Therefore, we get $$ \lim_{R\to\infty}\int_{\gamma_R}f(z)\,\mathrm{d}z=0 $$ where $\gamma_R=Re^{2\pi i[0,1]}$. This means that the sum of the residues must be $0$. Therefore, $$ 0=2\zeta(3)-2-\left(2-\frac{\pi^2}6\right)^2+\sum_{n=0}^\infty\left(\frac{2H_n}{(n+1)(n+2)^3}-\frac{4n+5}{(n+1)^2(n+2)^4}\right) $$ From there, he solves for $\sum\limits_{n=0}^\infty\frac{2H_n}{(n+1)(n+2)^3}$.

The Residue of $\boldsymbol{f(z)}$ at $\boldsymbol{n\in\mathbb{Z}}$ where $\boldsymbol{n\ge0}$

Consider $(\gamma+\psi(z))^2$: $$ \small\left(1+\color{#0000FF}{\frac1z}+\frac12+\color{#0000FF}{\frac1{z-1}}+\frac13+\color{#0000FF}{\frac1{z-2}}+\dots\right)\left(1+\color{#0000FF}{\frac1z}+\frac12+\color{#0000FF}{\frac1{z-1}}+\frac13+\color{#0000FF}{\frac1{z-2}}+\dots\right) $$ At $z=n$, the coefficient of $\frac1{(z-n)^2}$ is $1$.

At $z=n$, the coefficient of $\frac1{z-n}$ is $$ \small2\left(1+\color{#0000FF}{\frac1n}+\frac12+\color{#0000FF}{\frac1{n-1}}+\dots+\frac1n+\color{#0000FF}{1}+\frac1{n+1}+\color{#C00000}{0}+\frac1{n+2}-\color{#0000FF}{1}+\frac1{n+3}-\color{#0000FF}{\frac12}+\dots\right) $$ The factor of $2$ in front is because there is one $\frac1{z-n}$ in the left factor and one in the right. The red $0$ is a placeholder for the $\frac1{z-n}$ that was accounted for in the coefficient for $\frac1{(z-n)^2}$.

The blue $-1-\frac12-\dots$ cancels the black $1+\frac12+\dots$

The remainder is the blue $\frac1n+\frac1{n-1}+\frac1{n-2}+\dots+1=H_n$.

Thus, the coefficient of $\frac1{z-n}$ is $2H_n$.

The coefficient of $\frac1{z-n}$ in $f(z)$ is $$ \color{#0000FF}{2H_n}\frac1{(n+1)(n+2)^3}+\color{#0000FF}{1}\left[\frac{\mathrm{d}}{\mathrm{d}z}\frac1{(z+1)(z+2)^3}\right]_{z=n} $$ which is $$ \frac{2H_n}{(n+1)(n+2)^3}-\frac{4n+5}{(n+1)^2(n+2)^4} $$

robjohn
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  • This is a phenomenal answer. Do you mind if I ask a few questions?

    Also, how did you get this residues idea? :

    "Residue of $f(z)$ at $z=n\in\mathbb{Z}$ for $n\ge0$ is $$ \frac{2H_n}{(n+1)(n+2)^3}-\frac{4n+5}{(n+1)^2(n+2)^4} $$" How did you get this? I tried really hard, but I cant get the residues in form of a sum involving harmonic numbers.

    Also, I have always been interested.

    $|f(z)|\sim\frac{\log(|z|)^2}{|z|^3}$

    How can you say those are similar?

     – Amad27 Dec 28 '14 at 19:16
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    @Amad27: I have added a section on the computation of the residue of $f(z)$ at $z=n\in\mathbb{Z}$ for $n\ge0$. – robjohn Dec 31 '14 at 14:54
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    @Amad27: As long as we choose a path that stays more than $\frac13$ away from a negative integer, $\psi(z)\sim\log(z)$ for large $z$, we get that $\left|,\frac{(\gamma+\psi(-z))^2}{(z+1)(z+2)^3},\right| \sim\frac{\log(|z|)^2}{|z|^4}$. – robjohn Dec 31 '14 at 14:58
  • this is even better. Care for another question? How did you get that series representation for $f(z)$? If you can tell me that, I think I almost got this. =) – Amad27 Dec 31 '14 at 15:10
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    @Amad27: I don't give a series representation for $f(z)$. What is it you are asking for? – robjohn Dec 31 '14 at 15:26
  • Its not showing properly, it is the line below: consider $(\gamma + \psi(z))^2$ – Amad27 Dec 31 '14 at 15:30
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    That is just from the line which says $$\gamma+\psi(-z)=\sum_{k=0}^\infty\left(\frac1{k+1}+\frac1{z-k}\right)$$ This is a common formula for $\psi(z)$ that can be found on the Wikipedia page. – robjohn Dec 31 '14 at 15:33
  • right, sorry, next. How is the coefficient of $1/(z-n)^2$ at $z=n$ $=1$? $1/(z-n)^2$ doesnt exist at $z=n$? – Amad27 Dec 31 '14 at 15:35
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    @Amad27: There is one $\frac1{z-n}$ in the left factor and one $\frac1{z-n}$ in the right factor. Thus, in the product of the left and the right, we have $\frac1{(z-n)^2}$. – robjohn Dec 31 '14 at 17:59
  • Thanks. I understood that, but when you let $z=n$ it becomes undefined. So the coefficent cant be $1$ because the whole thing is undefined. Wait, Are you talking about a different $z$ like $z_1$ such that $z \ne n$? – Amad27 Dec 31 '14 at 18:24
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    @Amad27: yes. When talking about the residues and poles of a complex function at $z=\alpha$, we need to talk about $z$ in a neighborhood of $\alpha$. – robjohn Dec 31 '14 at 18:30
  • And following: "at $z=n$ the coefficient of $1/(z-n)$ is..." How did you get the following line? I cant seem to understand, how you get the coefficients? – Amad27 Dec 31 '14 at 18:40
  • Can we please chat? I am really really interested in this method, and you are helping me greatly, will you please chat? – Amad27 Dec 31 '14 at 18:43
  • Let us continue this discussion in chat. – Amad27 Dec 31 '14 at 18:51
  • What "coefficients," are you talking about? What are coefficients in this case? And why are you find them at $z=n$? Please, I am really requesting for your help. Not even a lot at once, just slight help. Pretty please...!! – Amad27 Jan 01 '15 at 20:17
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    We are talking about the coefficients of the Laurent Expansion about the point $z=n$. – robjohn Jan 01 '15 at 20:30
  • Where $n > 0$ or $n < 0$ also, how did you derive this coefficient? Wikipedia described a complicated line integral? – Amad27 Jan 01 '15 at 20:37
  • @Amad27: I've described, in detail, how to compute the coefficients for $n\ge0$. The singularities at $z=-1$ and $z=-2$ are simpler. – robjohn Jan 01 '15 at 20:50
  • Im sorry for being a bother, but I am not sure how to? I reread carefully, but I dont know how but I see what.... – Amad27 Jan 01 '15 at 20:54
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    @Amad27: You are not sure how to do what? Do you see how to compute the Laurent Expansion of $f(z)$ about $z=n$ for $n\in\mathbb{Z}$ and $n\ge0$? – robjohn Jan 02 '15 at 01:39
  • That is the issue, I am not sure for a general $n$. Will you perhaps show it for a simpler example? Thanks =) – Amad27 Jan 02 '15 at 21:52
  • Perhaps you can give me a link to where i can learn this? On wikipedia, the coefficient is a contour integral. I am not sure what method you are using. – Amad27 Jan 03 '15 at 09:26
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    Consider the case $n=1$. Look at the line following "Consider $(\gamma+\psi(z))^2$:" $$ \begin{align} &\left[\frac1{z-1}+\left(1+\color{#0000FF}{\frac1z}+\frac12 +\color{#C00000}{0} +\frac13+\color{#0000FF}{\frac1{z-2}}+\dots\right)\right]^2\ &=\frac1{(z-1)^2}+\frac2{z-1}\left(1+\color{#0000FF}{\frac1z}+\frac12 +\color{#C00000}{0} +\frac13+\color{#0000FF}{\frac1{z-2}}+\dots\right)\ &+\left(1+\color{#0000FF}{\frac1z}+\frac12+\color{#C00000}{0} +\frac13+\color{#0000FF}{\frac1{z-2}}+\dots\right)^2\ \end{align} $$ – robjohn Jan 03 '15 at 13:11
  • $$ \begin{align} &=\frac1{(z-1)^2}+\frac2{z-1}\left(1+\color{#0000FF}{\frac11}+\frac12 +\color{#C00000}{0} +\frac13+\color{#0000FF}{\frac1{-1}}+\dots+O(z-1)\right)\ &+\left(1+\color{#0000FF}{\frac11}+\frac12 +\color{#C00000}{0} +\frac13+\color{#0000FF}{\frac1{-1}}+\dots+O(z-1)\right)^2\ &=\frac1{(z-1)^2}+\frac{2\cdot\color{#0000FF}{\frac11}}{z-1}+O(1)\ \end{align} $$ – robjohn Jan 03 '15 at 13:11
  • I know $O$ means the continued pattern. But what would be $O(1)$ mean? I see what you did here, very nice. I see how you do it for one number. Now in a while I will try for the general $n$, until then can you please answer the above question? Thanks =) – Amad27 Jan 03 '15 at 13:52
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    @Amad27: No, $O$ is the Landau big-O. $O(1)$ represents a bounded function. – robjohn Jan 03 '15 at 15:17
  • What allows you to not let $z=1$ in $\frac{1}{(z-1)^2}$?? I tried for a general $n$: In the next comment – Amad27 Jan 04 '15 at 08:57
  • $$\frac{1}{(z-1)^2} + \frac{2}{z-1} \cdot \left( 1 + 1/n + 1/2 + 0 + 1/3 + 1/(n-2) \right) $$ Its not going correct as you can see... – Amad27 Jan 04 '15 at 09:00
  • @Amad27: why, for a general $n$, are you looking at powers of $z-1$? You want to expand in powers of $z-n$, not $z-1$. – robjohn Jan 04 '15 at 13:47
  • I am understanding this more. Why was it required to find the coefficient of $\frac{1}{(z-n)^2}$ ?? Also did you expand $(\psi(-z) + \gamma)^2)$ series form? and then find the coefficient? – Amad27 Jan 12 '15 at 08:45
  • I get the Harmonic sum part of the resiue at $z=n$ but I dont understand why $$\frac{d}{dz} \frac{1}{(z+1)(z+2)^3}$$ is also part of the residue. Care to help? Thanks! – Amad27 Jan 14 '15 at 18:56
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    @Amad27: Because of the $\frac1{(z-n)^2}$ from $(\gamma+\psi(z))^2$. We need the coefficient of the $(z-n)$ term from $\frac1{(z+1)(z+2)^3}$. – robjohn Jan 14 '15 at 21:04
  • Thanks for helping, okay so let see here. I am not sure. Why are we finding the coefficient of $\frac{1}{z-n}$ from the $\frac{1}{(z+1)(z+2)^3}$ term? – Amad27 Jan 15 '15 at 13:32
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    @Amad27: I did not say we want the coefficient of the $\frac1{z-n}$ term from $\frac{1}{(z+1)(z+2)^3}$; that is $0$ anyway. Because $(\gamma+\psi(z))^2$ has a singularity at $z=n$ whose highest order term is $\frac1{(z-n)^2}$ we need to know the coefficient of the $(z-n)$ term from $\frac{1}{(z+1)(z+2)^3}$ so that in the product of the two, we know the coefficient of the $\frac1{z-n}$ term. – robjohn Jan 15 '15 at 19:05
  • Is it also true that on any path that stays away from the negative integers, $\psi^{(m)} (z) \sim \frac{(-1)^{m-1}(m-1)!}{z^{m}}$ for large $z$? – Random Variable Feb 25 '20 at 15:27
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    As long as the path stays away from the negative real axis, say $\mathrm{Re}(z)\ge\lambda|z|$ for some $\lambda\gt-1$, and $|z|$ stays large. – robjohn Feb 25 '20 at 18:30
  • But then isn't $\psi(z) \sim \ln z$ for large $z$ also only true if the path stays away from the negative real axis? In your answer you seem to say that the path only needs to stay away from the negative integers, not the entire negative real axis. – Random Variable Feb 25 '20 at 19:34
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    Using the identity $\psi!\left(\frac12+z\right)-\psi!\left(\frac12-z\right)=\pi\tan(\pi z)$, we have $$ \psi!\left(-n+\tfrac12+iy\right)=\overbrace{\pi i\tanh(\pi y)}^\text{bounded}+\psi!\left(n+\tfrac12-iy\right) $$ so the estimate also works if the path crosses the real axis halfway between neighboring negative integers. – robjohn Feb 26 '20 at 13:16