What an Integral!!

Calculus Level 5

f ( x ) = e t x sinh 2 t d t \large f (x)=\int_{-\infty}^{\infty}e^{t-x\sinh^2t}dt

For f ( x ) f(x) as defined above, find the value of f ( π ) |f'(\pi)| , where f ( x ) f'(x) denotes the derivative of f ( x ) f(x) with respect to x x .


The answer is 0.15915.

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1 solution

Mark Hennings
Sep 14, 2019

The substitutions u = e t u = e^t and v = u 1 v = u^{-1} give f ( x ) = 0 e 1 4 x ( u u 1 ) 2 d u = 0 e 1 4 ( v v 1 ) 2 v 2 d v f(x) \; = \; \int_0^\infty e^{-\frac14x(u - u^{-1})^2}\,du \; = \; \int_0^\infty e^{-\frac14(v - v^{-1})^2}v^{-2}\,dv for x > 0 x > 0 , and hence f ( x ) = 1 2 0 e 1 4 x ( u u 1 ) 2 ( 1 + u 2 ) d u = 1 2 e 1 4 x w 2 d w = π x f(x) \; = \; \tfrac12\int_0^\infty e^{-\frac14x(u - u^{-1})^2}\,\big(1 + u^{-2}\big)\,du \; = \; \tfrac12\int_{-\infty}^\infty e^{-\frac14xw^2}\,dw \; = \; \sqrt{\tfrac{\pi}{x}} for x > 0 x > 0 , using the substitution w = u u 1 w = u - u^{-1} . Thus f ( x ) = π 2 x x x > 0 f'(x) \; = \; -\frac{\sqrt{\pi}}{2x\sqrt{x}} \hspace{2cm} x > 0 and so f ( π ) = 1 2 π f'(\pi) = -\tfrac{1}{2\pi} , making the answer 1 2 π \boxed{\tfrac{1}{2\pi}} .

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