[passer-by]

ข้อความเดิมเขียนโดยคุณ passer-by

26. Evaluate $$ \int_0^1 \int_0^1 \left\{\frac{1}{xy}\right\} \,\, dxdy $$

Note : {a} คือ fractional part ของ a เช่น {1.45}= 0.45

NOTE

Stieltjes constant : $$ \gamma_1 = \lim_{n \rightarrow \infty} \big( \sum_{k=1}^n \frac{\ln k}{k} - \frac{\ln^2 n}{2} \big) $$

BRIEF SOLUTION (Many details are omitted, please try by yourself for clearer understanding)

Let $ u= xy $ and $ v =\frac{y}{x}$

This will transform original integral into

$$ \frac{1}{2}\int_0^1 \int_u^\frac{1}{u} \frac{1}{v} \left\{\frac{1}{u}\right\} dvdu = \sum_{n=1}^{\infty} \int_\frac{1}{n+1}^\frac{1}{n} \ln(u) (n - \frac{1}{u}) \,\, du = \sum_{n=1}^{\infty} a_n $$

Consider partial sum of $a_n $,say, $S_N$

$$ S_N= -(\sum_{n=1}^N \frac{1}{n+1} - \ln(N+1))-(\sum_{n=1}^N \frac{\ln(n+1)}{n+1} - \frac{1}{2}\ln^2(N+1)) $$

Take limit $ n \rightarrow \infty $ ,and answer is $ 1- \gamma -\gamma_1$

NOTE : (1) You might not use reverse of integration as I hint above.

(2) Definition of $\gamma_1 $ is in note above