Prove the integral $\int_0^1 \frac{H_t}{t}dt=\sum_{k=1}^{\infty} \frac{\ln (1+\frac{1}{k})}{k}$

Question

By numerical results it follows that:

$$\int_0^1 \frac{H_t}{t}dt=\sum_{k=1}^{\infty} \frac{\ln (1+\frac{1}{k})}{k}=1.25774688694436963$$

Here $H_t$ is the harmonic number, which is the generalization of harmonic sum and has an integral representation:

$$H_t=\int_0^1 \frac{1-y^t}{1-y}~dy$$

---

If anyone has doubts about convergence, we have:

$$\lim_{t \to 0} \frac{H_t}{t} = \frac{\pi^2}{6}$$

Which would be another nice thing to prove, although I'm sure this proof is not hard to find.

---

It is also interesting that the related integral gives Euler-Mascheroni constant:

$$\int_0^1 H_t dt=\gamma$$

Answer 1

We apply $$ \frac{1-y^t}{1-y} = (1-y^t)(1+y+y^2+ \ldots)=(1-y^t)+y(1-y^t) +y^2(1-y^t)+\ldots $$ Each term gives after $y$-integration, $$ \frac t{t+1} + \frac{t}{2(t+2)}+ \frac t{3(t+3)} + \ldots $$ Then we divide these by $t$, $$ \frac 1{t+1} + \frac{1}{2(t+2)}+ \frac 1{3(t+3)} + \ldots $$ Taking integral with $t$ variable, we have the result.

Any interchange of integral and summation can be justified by Monotone Convergence Theorem.

Answer 2

Solution to the first part of this question was provided by @i707107.

---

For the second part, note that if we use the integral representation $H(t)=\int_0^1 \frac{1-y^t}{1-y}\,dy$, then we can write

$$\begin{align} \lim_{t\to 0} \frac{H(t)}{t}&=\lim_{t\to 0}\frac{\int_0^1 \frac{1-y^t}{1-y}\,dy}{t}\\\\ &=\lim_{t\to 0} \int_0^1 \frac{y^t \log(y)}{1-y}\,dy\\\\ &=-\int_0^1 \frac{\log(y)}{1-y}\,dy\\\\ &=-\int_0^1 \frac{\log(1-y)}{y}\,dy\\\\ &=\text{Li}_2(1)\\\\ &=\frac{\pi^2}{6} \end{align}$$

as was to be shown!

---

For the second part, we note that

$$\begin{align} \int_0^1 H(t)\,dt&=\int_0^1 \int_0^1 \frac{1-y^t}{1-y}\,dy\,dt\\\\ &=\int_0^1 \int_0^1 \sum_{n=0}^\infty (y^n-y^{n+t})\\\\ &=\sum_{n=0}^\infty \int_0^1 \left(\frac{1}{n+1}-\frac{1}{n+1+t}\right)\,dt\\\\ &=\sum_{n=0}^\infty \left(\frac{1}{n+1}-\log(n+2)+\log(n+1)\right)\\\\ &=\lim_{N\to \infty}\sum_{n=1}^N \left(\frac{1}{n}-\log(n+1)+\log(n)\right)\\\\ &=\lim_{N\to \infty}\left(\sum_{n=1}^N \frac1n -\log(N+1)\right)\\\\ &=\lim_{N\to \infty}\left(\sum_{n=1}^N \frac1n -\log(N)\right)\\\\ &=\gamma \end{align}$$

as was to be shown!

Answer 3

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Note that

• $\ds{ \sum_{k = 0}^{\infty}{1 \over \pars{k + \mu}\pars{k + \nu}} = {\Psi\pars{\mu} - \Psi\pars{\nu} \over \mu - \nu}\,,\qquad \pars{~\Psi:\ Digamma\ Function~}}$.
• $\ds{H_{z} = \Psi\pars{z + 1} - \Psi\pars{1}}$.

---

1. \begin{align} \color{#f00}{\int_{0}^{1}{H_{t} \over t}\,\dd t} & = \int_{0}^{1}{\Psi\pars{t + 1} - \Psi\pars{1} \over t}\,\dd t\qquad\qquad \\[5mm] & = \int_{0}^{1}\sum_{k = 0}^{\infty}{1 \over \pars{k + 1}\pars{k + t + 1}}\,\dd t = \sum_{k = 0}^{\infty}{1 \over k + 1}\,\ln\pars{k + 2 \over k + 1} = \color{#f00}{\sum_{k = 1}^{\infty}{1 \over k}\,\ln\pars{1 + {1 \over k}}} \end{align}

---

2. \begin{align} \color{#f00}{\lim_{t \to 0}{H_{t} \over t}} & = \lim_{t \to 0}{\Psi\pars{t + 1} - \Psi\pars{1} \over t} = \lim_{t \to 0}\sum_{k = 0}^{\infty}{1 \over \pars{k + 1}\pars{k + t + 1}} = \sum_{k = 1}^{\infty}{1 \over k^{2}} = \color{#f00}{\pi^{2} \over 6} \end{align}