Question Number 123261 by mnjuly1970 last updated on 24/Nov/20
$$\:\:\:\:\:\:\:\:\:\:…\:{nice}\:\:{calculus}… \\ $$$$\:\:\:\:{prove}\:\:{that}:: \\ $$$$ \\ $$$$\:\:\:\:\:\:\Omega=\int_{\mathbb{R}} {e}^{{x}−{sinh}^{\mathrm{2}} \left({x}\right)} {dx}=\sqrt{\pi} \\ $$
Answered by Olaf last updated on 24/Nov/20
$$\Omega\:=\:\int_{−\infty} ^{+\infty} {e}^{{x}} {e}^{−\mathrm{sinh}^{\mathrm{2}} {x}} {dx}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Let}\:{u}\:=\:−{x} \\ $$$$\Omega\:=\:\int_{+\infty} ^{−\infty} {e}^{−{x}} {e}^{−\mathrm{sinh}^{\mathrm{2}} {x}} \left(−{dx}\right)\:=\:\int_{−\infty} ^{+\infty} {e}^{−{x}} {e}^{−\mathrm{sinh}^{\mathrm{2}} {x}} {dx}\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\:: \\ $$$$\mathrm{2}\Omega\:=\:\int_{−\infty} ^{+\infty} \left({e}^{{x}} +{e}^{−{x}} \right){e}^{−\mathrm{sinh}^{\mathrm{2}} {x}} {dx} \\ $$$$\Rightarrow\:\Omega\:=\:\int_{−\infty} ^{+\infty} \mathrm{cosh}{xe}^{−\mathrm{sinh}^{\mathrm{2}} {x}} {dx} \\ $$$$\Rightarrow\:\Omega\:=\:\mathrm{2}\int_{\mathrm{0}} ^{+\infty} \mathrm{cosh}{xe}^{−\mathrm{sinh}^{\mathrm{2}} {x}} {dx} \\ $$$$\mathrm{Let}\:{u}\:=\:\mathrm{sinh}{x},\:{du}\:=\:\mathrm{cosh}{xdx} \\ $$$$\Omega\:=\:\mathrm{2}\int_{\mathrm{0}} ^{+\infty} {e}^{−{u}^{\mathrm{2}} } {du}\:=\:\mathrm{2}×\mathrm{erf}_{\infty} \:=\:\mathrm{2}×\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$$$\Omega\:=\:\sqrt{\pi} \\ $$
Commented by mnjuly1970 last updated on 24/Nov/20
$${bravo}\: \\ $$$${excellent}.{thank}\:{you}\:{master} \\ $$