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

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:

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

• Wait a minute, your integral doesn't converge. Aug 28, 2016 at 0:04
• @abnry, Mahtematica computes it no problem $$\text{NIntegrate}\left[\frac{H_t}{t},\{t,0,1\}\right]$$ Aug 28, 2016 at 0:06
• @abnry, Wolfram Alpha does too: wolframalpha.com/input/… Aug 28, 2016 at 0:09
• whoops, my bad, I thought your $H_t$ was constant. Aug 28, 2016 at 0:19
• To prove the second part see this: math.stackexchange.com/questions/4655283/… Mar 9 at 19:30

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.

• That's nice and clear, thank you. If you want, look at my profile for more integrals equal to this series Aug 28, 2016 at 0:30
• Thank you for letting me know. By the way, your limit $\pi^2/6$ also can be proved from my solution. Aug 28, 2016 at 0:33

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!

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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}