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Question Number 123261 by mnjuly1970 last updated on 24/Nov/20

          ... nice  calculus...      prove  that::          Ω=∫_R e^(x−sinh^2 (x)) dx=(√π)

$$\:\:\:\:\:\:\:\:\:\:...\:{nice}\:\:{calculus}... \\ $$$$\:\:\:\:{prove}\:\:{that}:: \\ $$$$ \\ $$$$\:\:\:\:\:\:\Omega=\int_{\mathbb{R}} {e}^{{x}−{sinh}^{\mathrm{2}} \left({x}\right)} {dx}=\sqrt{\pi} \\ $$

Answered by Olaf last updated on 24/Nov/20

Ω = ∫_(−∞) ^(+∞) e^x e^(−sinh^2 x) dx (1)  Let u = −x  Ω = ∫_(+∞) ^(−∞) e^(−x) e^(−sinh^2 x) (−dx) = ∫_(−∞) ^(+∞) e^(−x) e^(−sinh^2 x) dx (2)  (1)+(2) :  2Ω = ∫_(−∞) ^(+∞) (e^x +e^(−x) )e^(−sinh^2 x) dx  ⇒ Ω = ∫_(−∞) ^(+∞) coshxe^(−sinh^2 x) dx  ⇒ Ω = 2∫_0 ^(+∞) coshxe^(−sinh^2 x) dx  Let u = sinhx, du = coshxdx  Ω = 2∫_0 ^(+∞) e^(−u^2 ) du = 2×erf_∞  = 2×((√π)/2)  Ω = (√π)

$$\Omega\:=\:\int_{−\infty} ^{+\infty} {e}^{{x}} {e}^{−\mathrm{sinh}^{\mathrm{2}} {x}} {dx}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Let}\:{u}\:=\:−{x} \\ $$$$\Omega\:=\:\int_{+\infty} ^{−\infty} {e}^{−{x}} {e}^{−\mathrm{sinh}^{\mathrm{2}} {x}} \left(−{dx}\right)\:=\:\int_{−\infty} ^{+\infty} {e}^{−{x}} {e}^{−\mathrm{sinh}^{\mathrm{2}} {x}} {dx}\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\:: \\ $$$$\mathrm{2}\Omega\:=\:\int_{−\infty} ^{+\infty} \left({e}^{{x}} +{e}^{−{x}} \right){e}^{−\mathrm{sinh}^{\mathrm{2}} {x}} {dx} \\ $$$$\Rightarrow\:\Omega\:=\:\int_{−\infty} ^{+\infty} \mathrm{cosh}{xe}^{−\mathrm{sinh}^{\mathrm{2}} {x}} {dx} \\ $$$$\Rightarrow\:\Omega\:=\:\mathrm{2}\int_{\mathrm{0}} ^{+\infty} \mathrm{cosh}{xe}^{−\mathrm{sinh}^{\mathrm{2}} {x}} {dx} \\ $$$$\mathrm{Let}\:{u}\:=\:\mathrm{sinh}{x},\:{du}\:=\:\mathrm{cosh}{xdx} \\ $$$$\Omega\:=\:\mathrm{2}\int_{\mathrm{0}} ^{+\infty} {e}^{−{u}^{\mathrm{2}} } {du}\:=\:\mathrm{2}×\mathrm{erf}_{\infty} \:=\:\mathrm{2}×\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$$$\Omega\:=\:\sqrt{\pi} \\ $$

Commented by mnjuly1970 last updated on 24/Nov/20

bravo   excellent.thank you master

$${bravo}\: \\ $$$${excellent}.{thank}\:{you}\:{master} \\ $$

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