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## Yuriy S $(function() {$(".js-rank-badge").addSpinner().load("/users/rank?userId=64569"); }); My favorite number so far:

$$1.2577468869\dots= \gamma_{1}(1,0) - \gamma=$$

$$=\int_0^1 \frac{(1-x) \ln (1-x)}{x \ln x} \ {\rm d}x=$$ $$=\int_0^1 \ln (1-x) \ln \left(\ln \left(\frac{1}{x}\right)\right) \, dx-\gamma=$$

$$=\sum_{k=1}^{\infty} \frac{\ln (k+1)}{k(k+1)}=$$ $$=\sum_{k=1}^{\infty} \frac{\ln (1+\frac{1}{k})}{k}=$$ $$=\sum_{k = 2}^{\infty} \frac{(-1)^k \zeta(k)}{k-1}=$$

$$= \frac{\pi^2}{4}-1-4\int_{0}^{\infty} \frac{\arctan x}{1+x^{2}} \frac{dx}{e^{\pi x}+1}=$$ $$=\frac{\pi^2}{4}-1-4\int_{0}^{\pi/2} \frac{t~dt}{e^{\pi \tan t}+1} =$$

$$=\int_{1}^{\infty} \frac{\ln ([x]+1)}{x^2}dx=\int^{\infty}_0 \frac{\ln x+\Gamma (0,x) + \gamma}{e^x-1}~dx=$$

$$=\int^{\infty}_0 \int^1_0 \frac{e^{xt}-1}{t ~e^{xt}(e^x-1)} ~dt~dx=$$ $$=\int_0^1 \int_0^1 \frac{1-y^t}{t(1-y)}~dy~dt=\int_0^1 \frac{H_t}{t}dt$$

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