nice-calculus-prove-that-R-e-x-sinh-2-x-dx-pi-

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

\dots n i c e c a l c u l u s \dots

p r o v e t h a t ::

Ω = \int_{R} e^{x - {s i n h}^{2} (x)} d x = \sqrt{π}

Answered by Olaf last updated on 24/Nov/20

Ω = \int_{- \infty}^{+ \infty} e^{x} e^{- \sinh^{2} x} d x (1)

Let u = - x

Ω = \int_{+ \infty}^{- \infty} e^{- x} e^{- \sinh^{2} x} (- d x) = \int_{- \infty}^{+ \infty} e^{- x} e^{- \sinh^{2} x} d x (2)

(1) + (2) :

2 Ω = \int_{- \infty}^{+ \infty} (e^{x} + e^{- x}) e^{- \sinh^{2} x} d x

\Rightarrow Ω = \int_{- \infty}^{+ \infty} \cosh {x e}^{- \sinh^{2} x} d x

\Rightarrow Ω = 2 \int_{0}^{+ \infty} \cosh {x e}^{- \sinh^{2} x} d x

Let u = \sinh x, d u = \cosh x d x

Ω = 2 \int_{0}^{+ \infty} e^{- u^{2}} d u = 2 \times \erf_{\infty} = 2 \times \frac{\sqrt{π}}{2}

Ω = \sqrt{π}

Commented by mnjuly1970 last updated on 24/Nov/20

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