<s>[INST] Question about Riemann-Stieltjes problem with Euler Use Euler summation formula and integration by parts to show that
$$
\int_1^3 \left(\frac{[[x]]}{x^3}\right)dx = \frac{37}{72}
$$
Sorry I'm still learning how to use MathJax or how to put math formulas into the text...
Currently, I'm an undergraduate studying Math Analysis. Any help would be greatly appreaciated! Thanks for your time
 [/INST] Use the real function defined by $f(x)=\frac{-1}{2x^2}\Rightarrow f'(x)=\frac{1}{x^3}$
Euler summation formula implies:
$$\sum_{n=1}^3f(n)=\int_1^3f(x)dx+\int_1^3f'(x)(x-[|x|]-\frac{1}{2})dx+\frac{f(1)+f(3)}{2}$$
$$\sum_{n=1}^3(\frac{-1}{2n^2})=\int_1^3\frac{-1}{2x^2}dx+\int_1^3\frac{1}{x^3}(x-[|x|]-\frac{1}{2})dx-\frac{5}{18}$$
$$\sum_{n=1}^3(\frac{-1}{2n^2})=\int_1^3\frac{-1}{2x^2}dx+\int_1^3\frac{1}{x^2}dx-\int_1^3\frac{[|x|]}{x^3}dx-\frac{1}{2}\int_1^3\frac{1}{x^3}dx-\frac{5}{18}$$
Thus, you will get:
$$\int_1^3\frac{[|x|]}{x^3}dx=\int_1^3\frac{-1}{2x^2}dx+\int_1^3\frac{1}{x^2}dx-\frac{1}{2}\int_1^3\frac{1}{x^3}dx-\frac{5}{18}-\sum_{n=1}^3(\frac{-1}{2n^2})=\frac{37}{72}$$
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