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Question Number 157003 by MathSh last updated on 18/Oct/21

$$\mathrm{let}\mathbf{n}\in {\mathbb{Z}}^{+}$$$$\mathrm{shov}\mathrm{that}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{\sin \left({\mathrm{x}}^{-\mathbf{n}}\right)\ln \left(\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}=\frac{\pi \mathit{\gamma }}{{2\mathrm{n}}^2}$$$$\mathrm{where}\mathit{\gamma }\mathrm{is}\mathrm{the}\mathrm{Euler}-\mathrm{Mascheroni}\mathrm{constan}$$

Answered by mindispower last updated on 18/Oct/21

$${\int }_0^{\mathrm{\infty }}\frac{sin\left(x^n\right)ln\left(x\right)}{x}dx=\mathrm{\Delta }$$$$={\int }_0^{\mathrm{\infty }}\frac{sin\left(x^n\right)ln\left(x^n\right)}{{nx}^n}.x^{\mathit{n}-1}\mathit{d}\mathit{x}=\frac{1}{n^2}{\int }_0^{\mathrm{\infty }}\frac{sin\left(x^n\right)ln\left(x^n\right)}{x^n}{dx}^n$$$$=\frac{1}{n^2}{\int }_0^{\mathrm{\infty }}\frac{sin\left(t\right)}{t}ln\left(t\right)$$$$f\left(a\right)={\int }_0^{\mathrm{\infty }}sin\left(t\right)t^{a}dt=Im{\int }_0^{\mathrm{\infty }}{e}^{it}{t}^{a}dt$$$$=Im.{\int }_0^{i\mathrm{\infty }}{e}^{-t}{\left(it\right)}^a.idt$$$$={Imi}^{a+1}{\int }_0^{\mathrm{\infty }}{t}^a{e}^{-t}dt=sin\left(\frac{a+1}{2}\pi \right)\mathrm{\Gamma }(1+a)$$$$\mathrm{\Delta }=\frac{1}{n^2}{f}^{\prime }(-1)=\frac{\pi }{2n^2}\underset{a\to 0}{\lim }(cos\left(\frac{a}{2}\pi \right)\mathrm{\Gamma }\left(a\right)+{\mathrm{\Gamma }}^{\prime }\left(a\right)a)$$$${\mathrm{\Gamma }}^{\prime }\left(a\right)=\mathrm{\Psi }\left(a\right)\mathrm{\Gamma }\left(a\right)$$$$=\underset{a\to 0}{\lim }\frac{\pi }{2n^2}(\mathrm{\Gamma }\left(a\right)+\mathrm{\Psi }\left(a\right)\mathrm{\Gamma }\left(a\right)a)$$$$\mathrm{\Psi }(1+a)=\mathrm{\Psi }\left(a\right)+\frac{1}{a}\Rightarrow a\mathrm{\Psi }\left(a\right)=a\mathrm{\Psi }(1+a)-1$$$$=\underset{a\to 0}{\lim }\frac{\pi }{2n^2}(\mathrm{\Gamma }\left(a\right)+a\mathrm{\Psi }(1+a)\mathrm{\Gamma }\left(a\right)-\mathrm{\Gamma }\left(a\right))$$$$=\underset{a\to 0}{\lim }.\frac{\pi }{2n^2}\left(\mathrm{\Gamma }(1+a)\mathrm{\Psi }(1+a)\right)=\frac{\pi }{2n^2}\mathrm{\Psi }\left(1\right)=\frac{\pi \gamma }{2n^2}$$$$$$$$$$$$$$$$$$$$$$

Commented by MathSh last updated on 18/Oct/21

$$\mathrm{Perfect}\mathrm{dear}\mathrm{Ser},\mathrm{thank}\mathrm{you}$$

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