Prove-the-1-1-x-2-dx-Z-A-

Question Number 164163 by Zaynal last updated on 15/Jan/22

Prove the; ∫_(−∞) ^∞ (1/(1 + x^2 )) dx = 𝛑 ^({Z.A})

$$\mathrm{Prove}\:\mathrm{the}; \\ $$$$\int_{−\infty} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{1}\:+\:\boldsymbol{{x}}^{\mathrm{2}} }\:\boldsymbol{{dx}}\:=\:\boldsymbol{\pi} \\ $$$$\:^{\left\{\mathrm{Z}.\mathrm{A}\right\}} \\ $$

Answered by mathmax by abdo last updated on 15/Jan/22

∫_(−∞) ^(+∞) (dx/(1+x^2 ))=lim_(ξ→+∞) ∫_(−ξ) ^ξ (dx/(1+x^2 ))=lim_(ξ→+∞) [arctanx]_(−ξ) ^ξ =lim_(ξ→+∞) 2arctanξ =2.(π/2)=π or ∫_(−∞) ^(+∞) (dx/(1+x^2 ))=_(x=tanθ) ∫_(−(π/2)) ^(π/2) ((1+tan^2 θ)/(1+tan^2 θ))dθ =∫_(−(π/2)) ^(π/2) dθ =π

$$\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }=\mathrm{lim}_{\xi\rightarrow+\infty} \:\:\int_{−\xi} ^{\xi} \:\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }=\mathrm{lim}_{\xi\rightarrow+\infty} \left[\mathrm{arctanx}\right]_{−\xi} ^{\xi} \\ $$$$=\mathrm{lim}_{\xi\rightarrow+\infty} \mathrm{2arctan}\xi\:=\mathrm{2}.\frac{\pi}{\mathrm{2}}=\pi \\ $$$$\mathrm{or} \\ $$$$\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }=_{\mathrm{x}=\mathrm{tan}\theta} \:\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta}\mathrm{d}\theta\:=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{d}\theta\:=\pi \\ $$

Commented by Zaynal last updated on 15/Jan/22