integration - 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:

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

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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.

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

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

Answer

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.