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Question Number 79094 by mathmax by abdo last updated on 22/Jan/20

$$find{I}_{a,b}={\int }_0^{\mathrm{\infty }}\frac{sin\left(ax\right)sin\left(bx\right)}{x^2}dxwitha>0andb>0$$

Answered by mind is power last updated on 23/Jan/20

$$2sin\left(ax\right)sin\left(bx\right)=cos\left((a-b)x\right)-cos\left((a+b)x\right)$$ $$f\left(z\right)={\int }_0^{+\mathrm{\infty }}\frac{cos\left(zx\right)-1}{x^2}dx$$ $$z\in \mathbb{R}welldefind$$ $$inzerocos\left(zx\right)-1=-\frac{z^2{x}^2}{2}+o\left(x^3\right)$$ $$forz>1\frac{\mid cos\left(zx\right)-1\mid }{x^2}⩽\frac{2}{x^2},integrablin[1,+\mathrm{\infty }[$$ $$\mathrm{\forall }(z,x)\in IR\times {\mathbb{R}}^{\ast }(z,x)\to \frac{cos\left(zx\right)-1}{x^2}\in C_{\mathrm{\infty }}{C}_{1}iswhatweneed$$ $$\frac{\partial }{\partial z}\left(\frac{cos\left(zx\right)-1}{x^2}\right)=-\frac{sin\left(zx\right)}{x}\in C_1]0,+\mathrm{\infty }[$$ $${\int }_0^{+\mathrm{\infty }}-\frac{sin\left(zx\right)}{x}dx<\mathrm{\infty }\Rightarrow f\left(z\right)\in C_1$$ $$f^{\prime }\left(z\right)={\int }_0^{+\mathrm{\infty }}\frac{\partial }{\partial z}\left(\frac{cos\left(zx\right)-1}{x^2}\right)=-{\int }_0^{+\mathrm{\infty }}\frac{sin\left(zx\right)}{x}dx,u=zx$$ $$ifz>0$$ $$=-{\int }_0^{+\mathrm{\infty }}\frac{sin\left(u\right)}{u}du=-\frac{\pi }{2},z<0=\frac{\pi }{2}$$ $$\Rightarrow f^{\prime }\left(z\right)=-sign\left(z\right)\frac{\pi }{2}$$ $$f\left(z\right)=-\frac{zsign\left(z\right)\pi }{2}+c$$ $$f\left(0\right)=0\Rightarrow c=0$$ $$f\left(z\right)=\frac{-zsign\left(z\right)\pi }{2}$$ $$I_{a,b}={\int }_0^{+\mathrm{\infty }}\frac{sin\left(ax\right)sin\left(bx\right)}{x^2}dx$$ $$={\int }_0^{+\mathrm{\infty }}\frac{cos\left((a-b)x\right)-cos\left((a+b)x\right)}{2x^2}dx$$ $$={\int }_0^{+\mathrm{\infty }}\frac{cos\left((a-b)x\right)-1-(cos\left((a+b)x\right)-1)}{2x^2}dx$$ $$=\frac{1}{2}{\int }_0^{+\mathrm{\infty }}\frac{cos\left((a-b)x\right)-1}{x^2}-\frac{1}{2}{\int }_0^{+\mathrm{\infty }}\frac{cos\left((a+b)x\right)-1}{x^2}$$ $$=\frac{1}{2}f\left((a-b)\right)-\frac{1}{2}f(a+b))=-\frac{(a-b)sivn(a-b)\pi }{4}+\frac{(a+b)sign(a+b)\pi }{4}$$ $$=\frac{\pi }{4}((a+b)sign(a+b)-(a-b)sign(a-b))$$ $$$$ $$$$ $$$$ $$$$

Commented bymsup trace by abdo last updated on 23/Jan/20

$$thankyousir.$$

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