Question Number 129545 by mnjuly1970 last updated on 16/Jan/21
Answered by mindispower last updated on 16/Jan/21
![=∫_0 ^∞ ((sin^2 (x))/x^2 )−∫_0 ^∞ ((sin^2 (x))/(1+x^2 ))dx ∫_0 ^∞ ((sin^2 (x))/(1+x^2 ))dx=∫_0 ^∞ ((1−cos(2x))/(2(1+x^2 )))dx=(π/4)−(1/2)Re∫_0 ^∞ (e^(2ix) /(1+x^2 )) =(π/4)−((Re)/2).2iπ.(e^(−2) /(2i))=(π/4)−(π/(2e^2 )) ∫_0 ^∞ ((sin^2 (x))/x^2 )=[−((sin^2 (x))/x)]_0 ^∞ +∫_0 ^∞ ((sin(2x))/x)dx =(π/2) we get (π/4)(1+(2/e^2 ))](https://www.tinkutara.com/question/Q129548.png)
$$=\int_{\mathrm{0}} ^{\infty} \frac{{sin}^{\mathrm{2}} \left({x}\right)}{{x}^{\mathrm{2}} }−\int_{\mathrm{0}} ^{\infty} \frac{{sin}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}=\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}{Re}\int_{\mathrm{0}} ^{\infty} \frac{{e}^{\mathrm{2}{ix}} }{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=\frac{\pi}{\mathrm{4}}−\frac{{Re}}{\mathrm{2}}.\mathrm{2}{i}\pi.\frac{{e}^{−\mathrm{2}} }{\mathrm{2}{i}}=\frac{\pi}{\mathrm{4}}−\frac{\pi}{\mathrm{2}{e}^{\mathrm{2}} } \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}^{\mathrm{2}} \left({x}\right)}{{x}^{\mathrm{2}} }=\left[−\frac{{sin}^{\mathrm{2}} \left({x}\right)}{{x}}\right]_{\mathrm{0}} ^{\infty} +\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left(\mathrm{2}{x}\right)}{{x}}{dx} \\ $$$$=\frac{\pi}{\mathrm{2}} \\ $$$${we}\:{get}\:\frac{\pi}{\mathrm{4}}\left(\mathrm{1}+\frac{\mathrm{2}}{{e}^{\mathrm{2}} }\right) \\ $$
Commented by mnjuly1970 last updated on 16/Jan/21
$${god}\:{keep}\:{you}\:..{mr}\:{power}.. \\ $$
Answered by mathmax by abdo last updated on 16/Jan/21
![Φ=∫_0 ^∞ ((sin^2 x)/(x^2 +x^4 ))dx ⇒Φ=∫_0 ^∞ ((sin^2 x)/(x^2 (1+x^2 )))dx =∫_0 ^∞ ((1/x^2 )−(1/(x^2 +1)))sin^2 xdx =∫_0 ^∞ ((sin^2 x)/x^2 )dx−∫_0 ^∞ ((sin^2 x)/(x^2 +1))dx by parts ∫_0 ^∞ ((sin^2 x)/x^2 )dx =[−((sin^2 x)/x)]_0 ^∞ +∫_0 ^∞ (1/x)2sinx cosxdx =∫_0 ^∞ ((sin(2x))/x)dx =∫_0 ^∞ ((sin(2x))/(2x))d(2x)=(π/2) ∫_0 ^∞ ((sin^2 x)/(x^2 +1))dx =∫_0 ^∞ ((1−cos(2x))/(2(x^2 +1)))dx=(1/2)∫_0 ^∞ (dx/(x^2 +1))−(1/2)∫_0 ^∞ ((cos(2x))/(x^2 +1))dx =(π/4)−(1/4)∫_(−∞) ^(+∞) ((cos(2x))/(x^2 +1))dx and∫_(−∞) ^(+∞) ((cos(2x))/(x^(2 ) +1))dx=Re(∫_(−∞) ^(+∞) (e^(2ix) /(x^2 +1))dx) =Re(2iπ (e^(−2) /(2i))) =(π/e^2 ) ⇒Φ =(π/2)−((π/4)−(π/(4e^2 ))) =(π/4)+(π/(4e^2 )) ⇒Φ=(π/4)(1+(1/e^2 )).](https://www.tinkutara.com/question/Q129556.png)
$$\Phi=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{4}} }\mathrm{dx}\:\Rightarrow\Phi=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx}\:=\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right)\mathrm{sin}^{\mathrm{2}} \mathrm{xdx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}{\mathrm{x}^{\mathrm{2}} }\mathrm{dx}−\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx}\:\:\mathrm{by}\:\mathrm{parts} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}{\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:=\left[−\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}{\mathrm{x}}\right]_{\mathrm{0}} ^{\infty} +\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{x}}\mathrm{2sinx}\:\mathrm{cosxdx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}\left(\mathrm{2x}\right)}{\mathrm{x}}\mathrm{dx}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}\left(\mathrm{2x}\right)}{\mathrm{2x}}\mathrm{d}\left(\mathrm{2x}\right)=\frac{\pi}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)}\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx} \\ $$$$=\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}\int_{−\infty} ^{+\infty} \:\frac{\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx}\:\:\mathrm{and}\int_{−\infty} ^{+\infty} \:\frac{\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{x}^{\mathrm{2}\:} +\mathrm{1}}\mathrm{dx}=\mathrm{Re}\left(\int_{−\infty} ^{+\infty} \:\frac{\mathrm{e}^{\mathrm{2ix}} }{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx}\right) \\ $$$$=\mathrm{Re}\left(\mathrm{2i}\pi\:\frac{\mathrm{e}^{−\mathrm{2}} }{\mathrm{2i}}\right)\:\:=\frac{\pi}{\mathrm{e}^{\mathrm{2}} }\:\Rightarrow\Phi\:=\frac{\pi}{\mathrm{2}}−\left(\frac{\pi}{\mathrm{4}}−\frac{\pi}{\mathrm{4e}^{\mathrm{2}} }\right) \\ $$$$=\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{4e}^{\mathrm{2}} }\:\Rightarrow\Phi=\frac{\pi}{\mathrm{4}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{e}^{\mathrm{2}} }\right). \\ $$
Commented by mnjuly1970 last updated on 16/Jan/21
$${thanks}\:{alot}\:{mr}\:{max}… \\ $$
Commented by mathmax by abdo last updated on 16/Jan/21
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$