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Question Number 138967 by mnjuly1970 last updated on 20/Apr/21

$$\mathrm{\Theta }={\int }_0^{\mathrm{\infty }}\frac{sin\left(x\right).{log}^2\left(x\right)}{x}dx=\frac{\pi }{24}(12{\gamma }^2+{\pi }^2)....✓$$

Answered by Dwaipayan Shikari last updated on 20/Apr/21

$$\chi \left(\mu \right)={\int }_0^{\mathrm{\infty }}\frac{sinx}{x^{\mu }}dx=\frac{\pi }{2\mathrm{\Gamma }\left(\mu \right)sin\left(\frac{\pi \mu }{2}\right)}$$$${\chi }^{″}\left(1\right)={\int }_0^{\mathrm{\infty }}\frac{sinx}{x}{log}^2\left(x\right)dx=\frac{\pi }{2}(\frac{{\pi }^2}{12}+{\gamma }^2)$$

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