Evaluating $\int_1^{\pi/2}\frac{\sin(x)\ln(x)}{x^2}dx$

Question

The question goes:

Evaluate $\int_1^{\pi/2}\frac{\sin(x)\ln(x)}{x^2}dx$.

I have tried by parts, substitutions and other techniques I have learned from first year...but I didn't get any thing look good...

Answer 1

The only viable solution using first-year knowledge is a series solution. We start by replacing the sine function with its respective Taylor series. Then we can integrate term by term $$I=\int_1^{\frac{\pi}{2}}\frac{\ln{x}\sin{x}}{x^2}dx=\int_1^{\frac{\pi}{2}}\frac{\ln{x}}{x^2}\left(x+\sum_{k=1}^\infty\frac{(-1)^kx^{2k+1}}{(2k+1)!}\right)dx$$$$=\frac{1}{2}\ln^2\frac{\pi}{2}+\sum_{k=1}^\infty\frac{(-1)^k}{(2k+1)!}\int_1^{\frac{\pi}{2}}x^{2k-1}\ln{x}dx$$$$=\frac{1}{2}\ln^2\frac{\pi}{2}+\sum_{k=1}^\infty\frac{(-1)^k}{(2k+1)!}\left(\frac{\ln{\frac{\pi}{2}}}{2}\frac{(\pi^2/4)^k}{k}-\frac{1}{4}\frac{(\pi^2/4)^k}{k^2}+\frac{1}{4}\frac{1}{k^2}\right)$$$$=\frac{1}{2}\ln^2\frac{\pi}{2}+\frac{1}{2}\ln\frac{\pi}{2}\sum_{k=1}^\infty\frac{(-\pi^2/4)^k}{k(2k+1)!}-\frac{1}{4}\sum_{k=1}^\infty\frac{(-\pi^2/4)^k}{k^2(2k+1)!}+\frac{1}{4}\sum_{k=1}^\infty\frac{(-1)^k}{k^2(2k+1)!}$$ Without special functions, this is probably the closest you can get to a solution.

Answer 2

In the same spirit as @phi-rate

$$I=\int_1^{\pi/2}\frac{\sin(x)\log(x)}{x^2}dx$$

performing a series expansion around $x=1$ $$\frac{\sin(x)\log(x)}{x^2}=\sum_{n=1}^\infty \frac{a_n}{n!}\,(x-1)^n$$ This gives $$I=\sum_{n=1}^\infty\frac {a_n}{(n+1)!} \left(\frac{\pi-2}2\right)^{n+1}$$

Using $s=\sin(1)$ and $c=\cos(1)$, the first coefficients are $$\left( \begin{array}{cc} n & a_n \\ 1 & s \\ 2 & 2 c-5 s \\ 3 & 23 s-15 c \\ 4 & 100 c-124 s \\ 5 & 789 s-720 c \\ 6 & 5750 c-5793 s \\ 7 & 48243 s-50911 c \\ 8 & 497096 c-449552 s \\ 9 & 4635705 s-5318136 c \\ 10 & 61946698 c-52414485 s \\ 11 & 644829151 s-781045991 c \\ 12 & 10603366508 c-8575541148 s \\ \end{array} \right)$$

Using only the above terms leads to $I=\color{red}{0.0731}296$ while the exact value is $I=0.0731869$

Using twice more terms, $I=\color{red}{0.073186}859$ while the exact value is $0.073186903$

Answer 3

Using the two points Taylor approximation given by @Clerk in comments to this question $$\sin(x)\sim\left(\frac{5}{\pi ^7}-\frac{1}{2 \pi ^5}\right) x^4 (x-\pi )^4+\left(\frac{1}{6 \pi ^3}-\frac{2}{\pi ^5}\right) x^3 (x-\pi )^3+\frac{x^2 (x-\pi )^2}{\pi ^3}-\frac{x (x-\pi )}{\pi }$$ whose error is $7.5\times 10^{-5}\,z^5$ and the fact that $$\int_1^{\frac \pi 2}x^n \, \log(x)_,dx=\frac 1{(n+1)^2 }\left(1+\left(\frac{\pi}{2 }\right)^{n+1} \left((n+1) \log \left(\frac{\pi }{2}\right)-1\right)\right)$$ we obtain $$I=\int_1^{\pi/2}\frac{\sin(x)\log(x)}{x^2}dx=\color{red}{0.07318}539$$ while the exact value is $0.07318690$