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Question Number 173486 by mnjuly1970 last updated on 12/Jul/22

$$$$$$\mathit{\varphi }={\int }_0^{\mathrm{\infty }}\frac{cos\left(ax\right)-sin\left(bx\right)}{x^2}dx=?$$$$$$$$\mathit{\varphi }={\int }_0^{\mathrm{\infty }}\frac{1-2{sin}^2\left(\frac{ax}{2}\right)-(1-2{sin}^2\left(\frac{bx}{2}\right))}{x{}^2}dx$$$$=2\{{\int }_0^{\mathrm{\infty }}{\left(\frac{sin\left(\frac{bx}{2}\right)}{x}\right)}^{2}dx={\mathrm{\Theta }}_1\}-2\{{\int }_0^{\mathrm{\infty }}{\left(\frac{sin\left(\frac{ax}{2}\right)}{x}\right)}^{2}dx={\mathrm{\Theta }}_2\}$$$${\mathrm{\Theta }}_1\stackrel{\frac{bx}{2}=t}{=}\frac{b}{2}{\int }_0^{\mathrm{\infty }}\frac{{sin}^2\left(t\right)}{t^2}dt=\frac{\pi b}{4}$$$$similarly:{\mathrm{\Theta }}_2=\frac{\pi a}{4}\therefore \mathit{\varphi }=\frac{\pi }{2}(\mid b\mid -\mid a\mid )$$$${Dirichlet}^{\prime }sintegrals:{\int }_0^{\mathrm{\infty }}\frac{sin\left(x\right)}{x}dx=\frac{\pi }{2}={\int }_0^{\mathrm{\infty }}\frac{{sin}^2\left(x\right)}{x^2}dx$$

Commented by mokys last updated on 15/Jul/22

$$$$$$\mathit{\varphi }\left(\mathit{a}\right)={\int }_0^{\mathrm{\infty }}\frac{\mathit{c}\mathit{o}\mathit{s}\left(\mathit{a}\mathit{x}\right)}{{\mathit{x}}^2}\mathit{d}\mathit{x}\to {\mathit{\varphi }}^{{}^{\prime }}\left(\mathit{a}\right)=-{\int }_0^{\mathrm{\infty }}\frac{\mathit{s}\mathit{i}\mathit{n}\left(\mathit{a}\mathit{x}\right)}{\mathit{x}}\mathit{d}\mathit{x}$$$$\left(\mathit{a}\mathit{x}\right)=\mathit{k}\to \mathit{d}\mathit{x}=\frac{\mathit{d}\mathit{k}}{\mathit{a}}$$$$$$$${\mathit{\varphi }}^{{}^{\prime }}\left(\mathit{a}\right)=-{\int }_0^{\mathrm{\infty }}\frac{\mathit{s}\mathit{i}\mathit{n}\mathit{k}}{\mathit{k}}\mathit{d}\mathit{k}=-\frac{\mathit{\pi }}{2}$$$$$$$$\therefore \mathit{\varphi }\left(\mathit{a}\right)=-\frac{\mathit{a}\mathit{\pi }}{2}$$$$$$$$\mathit{\varphi }\left(\mathit{b}\right)={\int }_0^{\mathrm{\infty }}\frac{\mathit{s}\mathit{i}\mathit{n}\left(\mathit{b}\mathit{x}\right)}{{\mathit{x}}^2}\mathit{d}\mathit{x}$$$$$$$$\mathit{l}\mathit{e}\mathit{t}:\mathit{b}\mathit{x}=\mathit{m}\to \mathit{d}\mathit{x}=\frac{\mathit{d}\mathit{m}}{\mathit{b}}$$$$$$$$\mathit{\varphi }\left(\mathit{b}\right)=\mathit{b}{\int }_0^{\mathrm{\infty }}\frac{\mathit{s}\mathit{i}\mathit{n}\left(\mathit{m}\right)}{{\mathit{m}}^2}\mathit{d}\mathit{m}=\frac{\mathit{b}\mathit{\pi }}{2}$$$$$$$$\therefore \mathit{\varphi }(\mathit{a},\mathit{b})=\frac{\mathit{\pi }}{2}(\mid \mathit{b}\mid -\mid \mathit{a}\mid )$$$$$$$$\mathit{A}\mathit{l}\mathit{d}\mathit{o}\mathit{l}\mathit{a}\mathit{i}\mathit{m}\mathit{y}\mathit{m}\mathit{o}\mathit{h}\mathit{a}\mathit{m}\mathit{m}\mathit{a}\mathit{d}$$

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