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Question Number 142656 by qaz last updated on 03/Jun/21

$${\int }_{-\pi }^{\pi }\frac{\mathrm{xsin}\mathrm{x}}{1+{\mathrm{x}}^2}\mathrm{dx}=?$$

Answered by Ar Brandon last updated on 03/Jun/21

$$\xi \left(\mathrm{a}\right)={\int }_0^{\pi }\frac{\mathrm{xsin}\left(\mathrm{ax}\right)}{1+{\mathrm{x}}^2}\mathrm{dx}\Rightarrow {\xi }^{\prime }\left(\mathrm{a}\right)={\int }_0^{\pi }\frac{{\mathrm{x}}^2\cos \left(\mathrm{ax}\right)}{1+{\mathrm{x}}^2}\mathrm{dx}$$$${\xi }^{\prime }\left(\mathrm{a}\right)={\int }_0^{\pi }\cos \left(\mathrm{ax}\right)\mathrm{dx}-{\int }_0^{\pi }\frac{\cos \left(\mathrm{ax}\right)}{1+{\mathrm{x}}^2}\mathrm{dx}=-{\int }_0^{\pi }\frac{\cos \left(\mathrm{ax}\right)}{1+{\mathrm{x}}^2}\mathrm{dx}$$$${\xi }^{″}\left(\mathrm{a}\right)={\int }_0^{\pi }\frac{\mathrm{xsin}\left(\mathrm{ax}\right)}{1+{\mathrm{x}}^2}\mathrm{dx}=\xi \left(\mathrm{a}\right)\Rightarrow {\xi }^{″}\left(\mathrm{a}\right)-\xi \left(\mathrm{a}\right)=0$$$$\xi \left(\mathrm{a}\right)=\alpha {\mathrm{e}}^{\mathrm{a}}+\beta {\mathrm{e}}^{-\mathrm{a}},\xi \left(0\right)=\alpha +\beta =0...\mathrm{eqn}\left(1\right)$$$${\xi }^{\prime }\left(\mathrm{a}\right)=\alpha {\mathrm{e}}^{\mathrm{a}}-\beta {\mathrm{e}}^{-\mathrm{a}},{\xi }^{\prime }\left(0\right)=\alpha -\beta =\pi -{\tan }^{-1}\left(\pi \right)...\mathrm{eqn}\left(2\right)$$$$\mathrm{eqn}\left(1\right)\mathrm{\&}\mathrm{eqn}\left(2\right)\Rightarrow 2\alpha =\pi -{\tan }^{-1}\left(\pi \right),2\beta ={\tan }^{-1}\left(\pi \right)-\pi$$$$\Rightarrow \xi \left(\mathrm{a}\right)=\frac{1}{2}(\pi -{\tan }^{-1}\left(\pi \right))({\mathrm{e}}^{\mathrm{a}}-{\mathrm{e}}^{-\mathrm{a}})=(\pi -{\tan }^{-1}\left(\pi \right))\sinh \left(\mathrm{a}\right)$$$${\int }_{-\pi }^{\pi }\frac{\mathrm{xsinx}}{1+{\mathrm{x}}^2}\mathrm{dx}=2\xi \left(1\right)=2(\pi -{\tan }^{-1}\left(\pi \right))\sinh \left(1\right)$$

Commented by Dwaipayan Shikari last updated on 03/Jun/21

$$Nice!$$

Commented by Ar Brandon last updated on 03/Jun/21

$$\mathrm{Yes}\mathrm{bro}.\mathrm{But}\mathrm{final}\mathrm{answer}\mathrm{seems}\mathrm{not}\mathrm{to}$$$$\mathrm{be}\mathrm{correct}.\mathrm{Still}\mathrm{checking}...$$

Commented by qaz last updated on 03/Jun/21

$${\int }_{-\pi }^{\pi }\cos \left(\mathrm{ax}\right)\mathrm{dx}\ne 0$$

Commented by Ar Brandon last updated on 03/Jun/21

$$\mathrm{Hmmm}...$$$$\mathrm{Right}\mathrm{answer}\mathrm{is}\mathrm{approximately}$$$$1,682984232.\mathrm{Help}\mathrm{if}\mathrm{possible}.$$

Commented by mindispower last updated on 03/Jun/21

$$nicesir$$

Answered by mindispower last updated on 04/Jun/21

$${\int }_0^{\pi }\frac{2xsin\left(x\right)}{(x+i)(x-i)}dx$$$$=Re\left\{{\int }_0^{\pi }\frac{sin\left(x\right)}{x+i}dx\right\}$$$$=Re{\int }_i^{\pi +i}\frac{sin(t-i)}{t}dt$$$$=Re{\int }_i^{\pi +i}\frac{sin\left(t\right)cos\left(i\right)-cos\left(t\right)sin\left(i\right)}{t}$$$$=Re\{cos\left(i\right){\int }_i^{\pi +i}\frac{sin\left(t\right)}{t}dt-sin\left(i\right){\int }_i^{\pi +i}\frac{cos\left(t\right)}{t}dx\}$$$$=Re\{cos\left(i\right)(Si(\pi +i)-Si\left(i\right))-sin\left(i\right)(Ci\left(i\right)-Ci(i+\pi )\}$$$$cos\left(i\right)=ch\left(1\right),sin\left(i\right)=ish\left(1\right)$$$$Si={\int }_0^x\frac{sin\left(t\right)}{t},Ci={\int }_x^{\mathrm{\infty }}\frac{cos\left(t\right)}{t}dt$$$$=\frac{1}{2}ch\left(1\right)(si(\pi +i)+si(\pi -i)-si\left(i\right)-si(-i))$$$$-sh\left(1\right)(iCi\left(i\right)-iCi(i+\pi ))-sh(-iCi(-i)+iCi(\pi -i))$$$$$$

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