Question Number 219619 by Nicholas666 last updated on 29/Apr/25
$$ \\ $$$$\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \left({x}^{\mathrm{2}} +\mathrm{1}\right)^{−\mathrm{1}/\mathrm{2}} {dx} \\ $$$$ \\ $$
Answered by Ghisom last updated on 29/Apr/25
$$\mathrm{diverges} \\ $$
Answered by aleks041103 last updated on 29/Apr/25
![∫^ (1/( (√(1+x^2 )))) dx = −i∫(1/( (√(1−(ix)^2 ))))d(ix)= =−i arcsin(ix) = s sin(is)=ix=−((e^s −e^(−s) )/(2i))=i sinh(s) ⇒s=arcsinh(x) ⇒∫^ (1/( (√(1+x^2 )))) dx = arcsinh(x)+C ⇒∫_0 ^( ∞) (1/( (√(1+x^2 )))) dx = [arcsinh(x)]_0 ^∞ →∞ The intgral diverges.](https://www.tinkutara.com/question/Q219632.png)
$$\int^{\:} \frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:{dx}\:=\:−{i}\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\left({ix}\right)^{\mathrm{2}} }}{d}\left({ix}\right)= \\ $$$$=−{i}\:{arcsin}\left({ix}\right)\:=\:{s} \\ $$$${sin}\left({is}\right)={ix}=−\frac{{e}^{{s}} −{e}^{−{s}} }{\mathrm{2}{i}}={i}\:{sinh}\left({s}\right) \\ $$$$\Rightarrow{s}={arcsinh}\left({x}\right) \\ $$$$\Rightarrow\int^{\:} \frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:{dx}\:=\:{arcsinh}\left({x}\right)+{C} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\:\:\infty} \frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:{dx}\:=\:\left[{arcsinh}\left({x}\right)\right]_{\mathrm{0}} ^{\infty} \rightarrow\infty \\ $$$${The}\:{intgral}\:{diverges}. \\ $$
Commented by Nicholas666 last updated on 29/Apr/25
$${yes},{thank}\:{you}\:{sir} \\ $$