Question Number 131637 by rs4089 last updated on 07/Feb/21
![](https://www.tinkutara.com/question/18538.png)
Answered by Dwaipayan Shikari last updated on 07/Feb/21
![∫_(−∞) ^∞ ((sinx)/(x(x^2 +1)))dx=∫_(−∞) ^∞ ((sinx)/x)−((xsinx)/((x^2 +1)))dx=π−(π/e) I(α)=∫_(−∞) ^∞ ((cosαx)/((x^2 +1)))=πe^(−α) I′(α)=∫_(−∞) ^∞ ((xsin(αx))/(x^2 +1))dx=πe^(−α) I′(1)=(π/e)](https://www.tinkutara.com/question/Q131643.png)
$$\int_{−\infty} ^{\infty} \frac{{sinx}}{{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx}=\int_{−\infty} ^{\infty} \frac{{sinx}}{{x}}−\frac{{xsinx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx}=\pi−\frac{\pi}{{e}} \\ $$$${I}\left(\alpha\right)=\int_{−\infty} ^{\infty} \frac{{cos}\alpha{x}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)}=\pi{e}^{−\alpha} \\ $$$${I}'\left(\alpha\right)=\int_{−\infty} ^{\infty} \frac{{xsin}\left(\alpha{x}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}=\pi{e}^{−\alpha} \\ $$$${I}'\left(\mathrm{1}\right)=\frac{\pi}{{e}} \\ $$
Commented by Lordose last updated on 07/Feb/21
![why I′(α)=πe^(−α) and not −απe^(−α)](https://www.tinkutara.com/question/Q131655.png)
$$\mathrm{why}\:\mathrm{I}'\left(\alpha\right)=\pi\mathrm{e}^{−\alpha} \:\mathrm{and}\:\mathrm{not}\:−\alpha\pi\mathrm{e}^{−\alpha} \\ $$
Commented by Dwaipayan Shikari last updated on 07/Feb/21
![I(α)=πe^(−α) I′(α)=−πe^(−α) Also I′(α)=−∫_(−∞) ^∞ ((xsin(αx))/(x^2 +1)) So πe^(−α) =∫_0 ^∞ ((x sin(αx))/(x^2 +1))dx](https://www.tinkutara.com/question/Q131662.png)
$${I}\left(\alpha\right)=\pi{e}^{−\alpha} \\ $$$${I}'\left(\alpha\right)=−\pi{e}^{−\alpha} \\ $$$${Also}\:{I}'\left(\alpha\right)=−\int_{−\infty} ^{\infty} \frac{{xsin}\left(\alpha{x}\right)}{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$${So}\:\pi{e}^{−\alpha} =\int_{\mathrm{0}} ^{\infty} \frac{{x}\:{sin}\left(\alpha{x}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$
Answered by mathmax by abdo last updated on 07/Feb/21
![Φ=∫_(−∞) ^(+∞) ((sinx)/(x^3 +x))dx ⇒Φ =∫_(−∞) ^(+∞) ((sinx)/(x(x^2 +1)))dx =∫_(−∞) ^(+∞) sinx((1/x)−(x/(x^2 +1)))dx =∫_(−∞) ^(+∞) ((sinx)/x)dx−∫_(−∞) ^(+∞) ((xsinx)/(x^2 +1))dx =π−∫_(−∞) ^(+∞) ((xsinx)/(x^2 +1))dx =π−I I=Im(∫_(−∞) ^(+∞) ((xe^(ix) )/(x^2 +1))dx) but ∫_(−∞) ^(+∞) ((xe^(ix) )/(x^2 +1))dx =2iπRes(f,i) =2iπ.((ie^(−1) )/(2i)) =πie^(−1) ⇒I =(π/e) ⇒ Φ =π−(π/e)](https://www.tinkutara.com/question/Q131671.png)
$$\Phi=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{sinx}}{\mathrm{x}^{\mathrm{3}} +\mathrm{x}}\mathrm{dx}\:\Rightarrow\Phi\:=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{sinx}}{\mathrm{x}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{dx} \\ $$$$=\int_{−\infty} ^{+\infty} \:\mathrm{sinx}\left(\frac{\mathrm{1}}{\mathrm{x}}−\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right)\mathrm{dx}\:=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{sinx}}{\mathrm{x}}\mathrm{dx}−\int_{−\infty} ^{+\infty} \:\frac{\mathrm{xsinx}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx} \\ $$$$=\pi−\int_{−\infty} ^{+\infty} \:\frac{\mathrm{xsinx}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx}\:=\pi−\mathrm{I} \\ $$$$\mathrm{I}=\mathrm{Im}\left(\int_{−\infty} ^{+\infty} \:\frac{\mathrm{xe}^{\mathrm{ix}} }{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx}\right)\:\mathrm{but}\:\int_{−\infty} ^{+\infty} \:\frac{\mathrm{xe}^{\mathrm{ix}} }{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx} \\ $$$$=\mathrm{2i}\pi\mathrm{Res}\left(\mathrm{f},\mathrm{i}\right)\:=\mathrm{2i}\pi.\frac{\mathrm{ie}^{−\mathrm{1}} }{\mathrm{2i}}\:=\pi\mathrm{ie}^{−\mathrm{1}} \:\Rightarrow\mathrm{I}\:=\frac{\pi}{\mathrm{e}}\:\Rightarrow\:\Phi\:=\pi−\frac{\pi}{\mathrm{e}} \\ $$