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

∫_(−π) ^π ((xsin x)/(1+x^2 ))dx=?

$$\int_{−\pi} ^{\pi} \frac{\mathrm{xsin}\:\mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}=? \\ $$

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

ξ(a)=∫_0 ^π ((xsin(ax))/(1+x^2 ))dx⇒ξ′(a)=∫_0 ^π ((x^2 cos(ax))/(1+x^2 ))dx ξ′(a)=∫_0 ^π cos(ax)dx−∫_0 ^π ((cos(ax))/(1+x^2 ))dx=−∫_0 ^π ((cos(ax))/(1+x^2 ))dx ξ′′(a)=∫_0 ^π ((xsin(ax))/(1+x^2 ))dx=ξ(a)⇒ξ′′(a)−ξ(a)=0 ξ(a)=αe^a +βe^(−a) , ξ(0)=α+β=0...eqn(1) ξ′(a)=αe^a −βe^(−a) , ξ′(0)=α−β=π−tan^(−1) (π)...eqn(2) eqn(1) & eqn(2) ⇒2α=π−tan^(−1) (π), 2β=tan^(−1) (π)−π ⇒ξ(a)=(1/2)(π−tan^(−1) (π))(e^a −e^(−a) )=(π−tan^(−1) (π))sinh(a) ∫_(−π) ^π ((xsinx)/(1+x^2 ))dx=2ξ(1)=2(π−tan^(−1) (π))sinh(1)

$$\xi\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{xsin}\left(\mathrm{ax}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\Rightarrow\xi'\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{x}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{ax}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$\xi'\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\pi} \mathrm{cos}\left(\mathrm{ax}\right)\mathrm{dx}−\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{cos}\left(\mathrm{ax}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}=−\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{cos}\left(\mathrm{ax}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$\xi''\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{xsin}\left(\mathrm{ax}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}=\xi\left(\mathrm{a}\right)\Rightarrow\xi''\left(\mathrm{a}\right)−\xi\left(\mathrm{a}\right)=\mathrm{0} \\ $$$$\xi\left(\mathrm{a}\right)=\alpha\mathrm{e}^{\mathrm{a}} +\beta\mathrm{e}^{−\mathrm{a}} ,\:\xi\left(\mathrm{0}\right)=\alpha+\beta=\mathrm{0}...\mathrm{eqn}\left(\mathrm{1}\right) \\ $$$$\xi'\left(\mathrm{a}\right)=\alpha\mathrm{e}^{\mathrm{a}} −\beta\mathrm{e}^{−\mathrm{a}} ,\:\xi'\left(\mathrm{0}\right)=\alpha−\beta=\pi−\mathrm{tan}^{−\mathrm{1}} \left(\pi\right)...\mathrm{eqn}\left(\mathrm{2}\right) \\ $$$$\mathrm{eqn}\left(\mathrm{1}\right)\:\&\:\mathrm{eqn}\left(\mathrm{2}\right)\:\Rightarrow\mathrm{2}\alpha=\pi−\mathrm{tan}^{−\mathrm{1}} \left(\pi\right),\:\mathrm{2}\beta=\mathrm{tan}^{−\mathrm{1}} \left(\pi\right)−\pi \\ $$$$\Rightarrow\xi\left(\mathrm{a}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\pi−\mathrm{tan}^{−\mathrm{1}} \left(\pi\right)\right)\left(\mathrm{e}^{\mathrm{a}} −\mathrm{e}^{−\mathrm{a}} \right)=\left(\pi−\mathrm{tan}^{−\mathrm{1}} \left(\pi\right)\right)\mathrm{sinh}\left(\mathrm{a}\right) \\ $$$$\int_{−\pi} ^{\pi} \frac{\mathrm{xsinx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}=\mathrm{2}\xi\left(\mathrm{1}\right)=\mathrm{2}\left(\pi−\mathrm{tan}^{−\mathrm{1}} \left(\pi\right)\right)\mathrm{sinh}\left(\mathrm{1}\right) \\ $$

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

Nice!

$${Nice}! \\ $$

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

Yes bro. But final answer seems not to be correct. Still checking...

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

Commented by qaz last updated on 03/Jun/21

∫_(−π) ^π cos (ax)dx≠0

$$\int_{−\pi} ^{\pi} \mathrm{cos}\:\left(\mathrm{ax}\right)\mathrm{dx}\neq\mathrm{0} \\ $$

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

Hmmm... Right answer is approximately 1,682 984 232. Help if possible.

$$\mathrm{Hmmm}...\: \\ $$$$\mathrm{Right}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{approximately} \\ $$$$\mathrm{1},\mathrm{682}\:\mathrm{984}\:\mathrm{232}.\:\mathrm{Help}\:\mathrm{if}\:\mathrm{possible}. \\ $$

Commented by mindispower last updated on 03/Jun/21

nice sir

$${nice}\:{sir}\: \\ $$

Answered by mindispower last updated on 04/Jun/21

∫_0 ^π ((2xsin(x))/((x+i)(x−i)))dx =Re{∫_0 ^π ((sin(x))/(x+i))dx} =Re∫_i ^(π+i) ((sin(t−i))/t)dt =Re∫_i ^(π+i) ((sin(t)cos(i)−cos(t)sin(i))/t) =Re{cos(i)∫_i ^(π+i) ((sin(t))/t)dt−sin(i)∫_i ^(π+i) ((cos(t))/t)dx} =Re{cos(i)(Si(π+i)−Si(i))−sin(i)(Ci(i)−Ci(i+π)} cos(i)=ch(1),sin(i)=ish(1) Si=∫_0 ^x ((sin(t))/t),Ci=∫_x ^∞ ((cos(t))/t)dt =(1/2)ch(1)(si(π+i)+si(π−i)−si(i)−si(−i)) −sh(1)(iCi(i)−iCi(i+π))−sh(−iCi(−i)+iCi(π−i))

$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{2}{xsin}\left({x}\right)}{\left({x}+{i}\right)\left({x}−{i}\right)}{dx} \\ $$$$={Re}\left\{\int_{\mathrm{0}} ^{\pi} \frac{{sin}\left({x}\right)}{{x}+{i}}{dx}\right\} \\ $$$$={Re}\int_{{i}} ^{\pi+{i}} \frac{{sin}\left({t}−{i}\right)}{{t}}{dt} \\ $$$$={Re}\int_{{i}} ^{\pi+{i}} \frac{{sin}\left({t}\right){cos}\left({i}\right)−{cos}\left({t}\right){sin}\left({i}\right)}{{t}} \\ $$$$={Re}\left\{{cos}\left({i}\right)\int_{{i}} ^{\pi+{i}} \frac{{sin}\left({t}\right)}{{t}}{dt}−{sin}\left({i}\right)\int_{{i}} ^{\pi+{i}} \frac{{cos}\left({t}\right)}{{t}}{dx}\right\} \\ $$$$={Re}\left\{{cos}\left({i}\right)\left({Si}\left(\pi+{i}\right)−{Si}\left({i}\right)\right)−{sin}\left({i}\right)\left({Ci}\left({i}\right)−{Ci}\left({i}+\pi\right)\right\}\right. \\ $$$${cos}\left({i}\right)={ch}\left(\mathrm{1}\right),{sin}\left({i}\right)={ish}\left(\mathrm{1}\right) \\ $$$${Si}=\int_{\mathrm{0}} ^{{x}} \frac{{sin}\left({t}\right)}{{t}},{Ci}=\int_{{x}} ^{\infty} \frac{{cos}\left({t}\right)}{{t}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ch}\left(\mathrm{1}\right)\left({si}\left(\pi+{i}\right)+{si}\left(\pi−{i}\right)−{si}\left({i}\right)−{si}\left(−{i}\right)\right) \\ $$$$−{sh}\left(\mathrm{1}\right)\left({iCi}\left({i}\right)−{iCi}\left({i}+\pi\right)\right)−{sh}\left(−{iCi}\left(−{i}\right)+{iCi}\left(\pi−{i}\right)\right) \\ $$$$ \\ $$

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