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:
$$\int_0^1 H_t dt=\gamma$$
 A: 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. 
A: 
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!
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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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}
