calculate-0-e-x-sin-x-2-x-2-dx-

Question Number 76782 by mathmax by abdo last updated on 30/Dec/19

c a l c u l a t e \int_{0}^{\infty} e^{- x} \frac{s i n (x^{2})}{x^{2}} d x

Answered by mind is power last updated on 04/Apr/20

A = \int_{0}^{+ \infty} e^{- x} \frac{s i n (x^{2})}{x^{2}} d x

s i n (x^{2}) = \sum_{n ⩾ 0} \frac{{(- 1)}^{n} . x^{4 n + 2}}{(2 n + 1)!}

A = \int_{0}^{+ \infty} \sum_{n ⩾ 0} \frac{{(- 1)}^{n} x^{4 n} e^{- x} d x}{(2 n + 1)!}

A = \sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{(2 n + 1)!} \int_{0}^{+ \infty} x^{4 n} e^{- x} d x

= \sum_{n ⩾ 0} \frac{{(- 1)}^{n} Γ (4 n + 1)}{(2 n + 1)!} = \sum_{n ⩾ 0} \frac{{(- 1)}^{n} . (4 n)!}{(2 n + 1)!}

= \sum_{n ⩾ 0} . \frac{{(- 1)}^{n} . \underset{k = 1}{\prod^{n}} (4 k) . \underset{k = 0}{\prod^{n - 1}} (4 k + 1) . \underset{k = 0}{\prod^{n - 1}} (4 k + 2) . \underset{k = 0}{\prod^{n - 1}} (4 k + 3)}{. \underset{k = 0}{\prod^{n - 1}} (2 k + 1) . \underset{k = 0}{\prod^{n - 1}} (2 k + 2)}

= \sum_{n ⩾ 0} \frac{{(- 1)}^{n} . 4^{4 n} . n! . \underset{k = 0}{\prod^{n - 1}} (k + \frac{1}{4}) . \underset{k = 0}{\prod^{n - 1}} (\frac{1}{2} + k) . \underset{k = 0}{\prod^{n - 1}} (k + \frac{3}{4})}{2^{2 n} \underset{k = 0}{\prod^{n - 1}} (\frac{1}{2} + k) . n!}

= \sum_{n ⩾ 0} \frac{. \underset{k = 0}{\prod^{n - 1}} (1 + k) . \underset{k = 0}{\prod^{n - 1}} (\frac{1}{4} + k) . \underset{k = 0}{\prod^{n - 1}} (\frac{3}{4} + k)}{\underset{k = 0}{\prod^{n - 1}} (\frac{1}{2} + k) .} . \frac{{(- 4^{3})}^{n}}{n!}

=_{3} F_{1} (1, \frac{1}{4}, \frac{3}{4}; \frac{1}{2}; - 4^{3})