Question Number 26631 by jkssm1857@gmail.com last updated on 27/Dec/17
$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$ \\ $$
Commented by abdo imad last updated on 27/Dec/17
![its a divergent integral let put Iε= ∫_ε ^∝ (dx/x^2 ) I(ε)= [−(1/x) ]_(x=ε) ^∝ = (1/ε) and lim_(ε_(ε>0) −>0) I(ε)=+∝](Q26671.png)
$${its}\:{a}\:{divergent}\:{integral}\:{let}\:{put}\:{I}\varepsilon=\:\int_{\varepsilon} ^{\propto} \frac{{dx}}{{x}^{\mathrm{2}} } \\ $$$${I}\left(\varepsilon\right)=\:\left[−\frac{\mathrm{1}}{{x}}\:\right]_{{x}=\varepsilon} ^{\propto} =\:\:\frac{\mathrm{1}}{\varepsilon}\:\:\:\:\:\:{and}\:\:{lim}_{\varepsilon_{\varepsilon>\mathrm{0}} −>\mathrm{0}} {I}\left(\varepsilon\right)=+\propto \\ $$