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July 6

Convergence and Closed Form Expression

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{b(x)}{B(x)}dx=?,b(x)=\sum _{n=1}^{\mathrm{\infty }}\frac{n^x}{n^n},B(x)=\sum _{n=1}^{\mathrm{\infty }}\frac{n^x}{n!}$

Observations:

1. B(x) is e times a Bell number.

2. If the numerator of each function would be xⁿ instead of n^x, and each sum would start at 0 instead of 1, then the value of our integral would become $\sum _{n=0}^{\mathrm{\infty }}\frac{n!}{n^n},$ where lim nⁿ → 1 when n → 0.

— 79.113.210.135 (talk) 20:45, 6 July 2013 (UTC)

You could try to use Ramanujan's master theorem. Count Iblis (talk) 21:40, 6 July 2013 (UTC)

In that case, s = 1 and $f(x)=\frac{b(x)}{B(x)},$ whose Taylor expansion is absolutely catastrophic... 79.113.210.135 (talk) 23:11, 6 July 2013 (UTC)

To make matters even more complicated than they already are, integration by parts would seem to suggest that the integral is divergent if any of the two sums starts from n = 1, despite numerical data indicating the contrary... The question now would be whether starting at n = 2 would really make that much of a difference anyway... 79.113.225.68 (talk) 13:58, 7 July 2013 (UTC)