Need Help with an Integral Formula for Alternating Sum of Reciprocals of Logarithms I was curious as to the limit of $\sum_{2}^{\infty} \frac{(-1)^n}{\ln(n)}$, and eventually found the sequence of its digits on the OEIS, sequence A099769. On there, the expression:
$$\frac{1}{2\ln(2)} + 8\int_{0}^{\infty} \frac{\arctan(x)}{\ln(4+4x^2)^2+4\arctan(x)^2\sinh(2\pi x)}dx$$
Is given as being equal to the relevant sum:
$$1/(2\log(2)) + 8 \int_{x=0} ^{\infty} \frac{\arctan(x)}{((\log(4+4*x^2)^2+4\arctan(x)^2)*\sinh(2\pi x)} dx$$
-Iaroslav V. Blagouchine, May 11 2015
However, computing this expression leaves me with a decimal value of 1.01756347165, whereas the accepted value of the series begins 0.924299897222.
Am I missing something? As far as I can tell, this expression does not appear equal to the constant.

You missed some parentheses. The integral, according to Blagouchine, should be
$$ \frac{1}{2 \ln \! \left(2\right)}+8 {{\int}}_{0}^{\infty}\frac{\arctan \! \left(x \right)}{\left(\ln \! \left(4+4 x^{2}\right)^{2}+4 \arctan \! \left(x \right)^{2}\right) \sinh \! \left(2 \pi  x \right)}{d}x 
$$
I don't know how Blagouchine gets that, but the numerical value is
approximately $0.9242998972229386$ according to Maple.
