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Question Number 52515 by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19

$$1){\int }_0^{\mathrm{\infty }}\frac{{tan}^{-1}\left(ax\right)-{tan}^{-1}\left(bx\right)}{x}dx$$$$2){\int }_0^{\mathrm{\infty }}\frac{e^{-x}sinx}{x}dx$$

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19

$$thankyousir...$$

Commented by maxmathsup by imad last updated on 09/Jan/19

$$1)letI\left(\xi \right)={\int }_0^{\xi }\frac{arctan\left(ax\right)-arctan\left(bx\right)}{x}dxwehave$$$${\int }_0^{\mathrm{\infty }}\frac{arctan\left(ax\right)-arctan\left(bx\right)}{x}dx={lim}_{\xi \to +\mathrm{\infty }}I\left(\xi \right)but$$$$I\left(\xi \right)={\int }_0^{\xi }\frac{arctan\left(ax\right)}{x}dx-{\int }_0^{\xi }\frac{arctan\left(bx\right)}{x}dxbut$$$${\int }_0^{\xi }\frac{arctan\left(ax\right)}{x}dx{=}_{ax=t}{\int }_0^{a\xi }\frac{arctan\left(t\right)}{\frac{t}{a}}\frac{dt}{a}={\int }_0^{a\xi }\frac{arctan\left(t\right)}{t}dtalso$$$${\int }_0^{\xi }\frac{arctan\left(bx\right)}{x}dx={\int }_0^{b\xi }\frac{arctan\left(t\right)}{t}dt\Rightarrow I\left(\xi \right)={\int }_0^{a\xi }\frac{arctan\left(t\right)}{t}dt-{\int }_0^{b\xi }\frac{arctan\left(t\right)}{t}dt$$$$={\int }_{b\xi }^{a\xi }\frac{arctan\left(t\right)}{t}dtbut\mathrm{\exists }c\in ]b\xi ,a\xi [/I\left(\xi \right)=arctan\left(\xi \right){\int }_{b\xi }^{a\xi }\frac{dt}{t}$$$$=arctan\left(\xi \right)ln(\mid \frac{a}{b}\mid )\to \frac{\pi }{2}ln(\mid \frac{a}{b}\mid )(\xi \to +\mathrm{\infty })\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{arctan\left(ax\right)-arctan\left(bx\right)}{x}dx=\frac{\pi }{2}ln\mid \frac{a}{b}\mid .$$

Commented by Abdo msup. last updated on 09/Jan/19

$$youarewelcome..$$

Answered by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19

$$1)Thereisaformula...Frullaniintregal...$$$$i)iff\left(x\right)iscontinuousfunctionofxfor\mathrm{\infty }>x⩾0$$$$ii)\underset{x\to 0}{\lim }f\left(x\right)=A\underset{x\to \mathrm{\infty }}{\lim }f\left(x\right)=B$$$$then{\int }_0^{\mathrm{\infty }}\frac{f\left(ax\right)-f\left(bx\right)}{x}dx=(A-B)ln\frac{b}{a}$$$$now$$$$heref\left(x\right)={tan}^{-1}\left(x\right)$$$$\underset{x\to 0}{\lim }{tan}^{-1}x=0\leftarrow A$$$$\underset{x\to \mathrm{\infty }}{\lim }{tan}^{-1}\left(x\right)=\frac{\pi }{2}\leftarrow B$$$$so{\int }_0^{\mathrm{\infty }}\frac{{tan}^{-1}\left(ax\right)-{tan}^{-1}\left(bx\right)}{x}\left[formula(A-B)ln\left(\frac{b}{a}\right)\right]$$$$=(0-\frac{\pi }{2})ln\left(\frac{b}{a}\right)$$$$=\frac{\pi }{2}ln\left(\frac{a}{b}\right)$$

Answered by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19

$$2){\int }^{\mathrm{\infty }}\frac{e^{-x}sinx}{x}dx$$$$=I(a,b)={\int }_0^{\mathrm{\infty }}\frac{e^{-ax}sinbx}{x}dx$$$$\frac{dI}{db}={\int }_0^{\mathrm{\infty }}\frac{e^{-ax}}{x}\times \frac{\partial }{\partial b}\left(sinbx\right)dx$$$$={\int }_0^{\mathrm{\infty }}\frac{e^{-ax}}{x}\times cosbx\times xdx$$$$={\int }_0^{\mathrm{\infty }}{e}^{-ax}cosbxdx$$$$nowformula$$$${\int }_0^{\mathrm{\infty }}{e}^{-ax}sinbxdx=\frac{b}{a^2+b^2}$$$${\int }_0^{\mathrm{\infty }}{e}^{-ax}cosbx=\frac{a}{a^2+b^2}$$$$so{\int }_0^{\mathrm{\infty }}{e}^{-ax}cosbxdx=\frac{a}{a^2+b^2}$$$$\frac{dI}{db}=\frac{a}{a^2+b^2}$$$$\int dI=a\int \frac{db}{a^2+b^2}$$$$I=a\times \frac{1}{a}{tan}^{-1}\left(\frac{b}{a}\right)+c$$$$nowI(a,b)={\int }_0^{\mathrm{\infty }}\frac{e^{-ax}sinbx}{x}dx$$$$putb=0I=0soconstantvaluec=0$$$$soI={tan}^{-1}\left(\frac{b}{a}\right)$$$$nowputa=1b=1$$$$I={tan}^{-1}\left(1\right)=\frac{\pi }{4}Ans$$

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