$$\int^{\infty}_0 \frac{\log t+\Gamma (0,t) + \gamma}{e^t-1}~dt=\sum_{k = 2}^{\infty} \frac{(-1)^k \zeta(k)}{k-1}=1.2577468869$$

Here $\gamma$ -Euler constant, $\Gamma (0,t)$ - incomplete gamma function:

$$\Gamma (0,t)=\int_t^{\infty} \frac{e^{-p}}{p} dp=\cfrac{\exp(-t)}{t+1-\cfrac{1}{t+3-\cfrac{4}{t+5-\cfrac{9}{t+7-\cdots}}}}$$

What is so special about this integral? None of the terms converge on their own, according to WolframAlpha:

$$\int^{\infty}_0 \frac{\log t}{e^t-1} ~dt \to \infty$$

$$\int^{\infty}_0 \frac{\Gamma(0,t)}{e^t-1} ~dt \to \infty$$

$$\int^{\infty}_0 \frac{\gamma}{e^t-1} ~dt \to \infty$$

Moreover, no two terms together converge (again, according to WolframAlpha).

This integral is equal to another integral and several infinite series:

$$\int^{\infty}_0 \frac{\log t+\Gamma (0,t) + \gamma}{e^t-1}~dt=$$

$$=\frac{\pi^2}{4}-1-4\int_{0}^{\infty} \frac{\arctan x}{1+x^{2}} \frac{dx}{e^{\pi x}+1}=\sum_{k=1}^{\infty} \frac{\ln (1+\frac{1}{k})}{k}=\sum_{k = 2}^{\infty} \frac{(-1)^k \zeta(k)}{k-1}$$

See this answer and this answer for clarification.

And now proof. See this question for the proof of the following integral:

$$\int^1_0 \frac{\log^n x}{1-x}dx=(-1)^n~ n!~ \zeta(n+1)$$

Now we change the variable under the integral:

$$\log x=-t$$

$$\int^{\infty}_0 \frac{(-1)^n t^n \exp(-t)}{1-\exp(-t)}dt=(-1)^n~ n!~ \zeta(n+1)$$

Now we divide both sides by $n!~n$ and sum from $n=1$ to $\infty$:

$$\sum_{n=1}^{\infty} \frac{(-1)^n t^n }{n!n}=-\log t-\Gamma (0,t) - \gamma$$

See WolframAlpha.

Finally, we can write:

$$-\int^{\infty}_0 \frac{\exp(-t)}{1-\exp(-t)} (\log t+\Gamma (0,t) + \gamma)~dt=\sum_{n = 1}^{\infty} \frac{(-1)^n \zeta(n+1)}{n}$$

$$\int^{\infty}_0 \frac{\log t+\Gamma (0,t) + \gamma}{e^t-1}~dt=\sum_{n = 2}^{\infty} \frac{(-1)^n \zeta(n)}{n-1}$$

The rest of the equalitites, including the numerical value of the integral follows from this one.

Is it a known integral? Have you seen it before? Some reference of other proofs would be nice, since I don't know if anyone tried to prove it the same way

  • Yes, yes and maybe, one cannot really know when and if some integral is "more interesting" or "more useful" than some other. – Von Neumann Mar 3 '16 at 9:55
  • Can you provide some reference? – Yuriy S Mar 3 '16 at 10:00

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