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Question Number 150533 by mnjuly1970 last updated on 13/Aug/21

          solve...    I:= ∫_(−∞) ^( ∞) e^( x−n.sinh^( 2) (x)) dx =^(??) (√(π/n))         ...m.n...

$$\:\:\:\: \\ $$$$\:\:\:\:\mathrm{solve}... \\ $$$$\:\:\mathrm{I}:=\:\int_{−\infty} ^{\:\infty} {e}^{\:{x}−{n}.{sinh}^{\:\mathrm{2}} \left({x}\right)} {dx}\:\overset{??} {=}\sqrt{\frac{\pi}{{n}}} \\ $$$$\:\:\:\:\:\:\:...{m}.{n}... \\ $$

Answered by Kamel last updated on 13/Aug/21

  Ω=∫_(−∞) ^(+∞) e^(x−nsinh^2 (x)) dx=∫_(−∞) ^(+∞) e^(−x−nsinh^2 (x)) dx  Ω=∫_(−∞) ^(+∞) e^(−nsinh^2 (x)) (((e^x +e^(−x) )/2))dx=^(u=sinh(x)) ∫_(−∞) ^(+∞) e^(−nu^2 ) du     =(√(π/n))

$$ \\ $$$$\Omega=\int_{−\infty} ^{+\infty} {e}^{{x}−{nsinh}^{\mathrm{2}} \left({x}\right)} {dx}=\int_{−\infty} ^{+\infty} {e}^{−{x}−{nsinh}^{\mathrm{2}} \left({x}\right)} {dx} \\ $$$$\Omega=\int_{−\infty} ^{+\infty} {e}^{−{nsinh}^{\mathrm{2}} \left({x}\right)} \left(\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}}\right){dx}\overset{{u}={sinh}\left({x}\right)} {=}\int_{−\infty} ^{+\infty} {e}^{−{nu}^{\mathrm{2}} } {du} \\ $$$$\:\:\:=\sqrt{\frac{\pi}{{n}}} \\ $$

Commented by mnjuly1970 last updated on 13/Aug/21

excellent master  ...very nice..

$${excellent}\:{master}\:\:...{very}\:{nice}.. \\ $$

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