nice-calculus-prove-that-I-0-1-e-x-sin-2-x-x-3-2-2pi-1-2-1-m-n-

Question Number 154409 by mnjuly1970 last updated on 18/Sep/21

n i c e c a l c u l u s . .

p r o v e t h a t :

I := \int_{0}^{\infty} \frac{(1 + e^{- x}) {s i n}^{2} (x)}{x^{\frac{3}{2}}} = \sqrt{2 π} (1 + \sqrt{\sqrt{2} - 1})

m . n

Answered by Kamel last updated on 18/Sep/21

I := \int_{0}^{\infty} \frac{(1 + e^{- x}) {s i n}^{2} (x)}{x^{\frac{3}{2}}}

= 2 \int_{0}^{1} \int_{0}^{+ \infty} x^{- \frac{1}{2}} (1 + e^{- x}) s i n (2 a x) d x d a

= 2 \int_{0}^{1} (\sqrt{2} a^{- \frac{1}{2}} \int_{0}^{+ \infty} s i n (u^{2}) d u + \frac{1}{2 i} \int_{0}^{+ \infty} (e^{- (1 - 2 a i) x^{2}} - e^{- (1 + 2 a i) x^{2}}) d x) d a

= 2 \int_{0}^{1} (2^{- \frac{1}{2}} a^{- \frac{1}{2}} \frac{\sqrt{π}}{2 \sqrt{2}} + \frac{\sqrt{π}}{4 i} (\frac{1}{\sqrt{1 - 2 a i}} - \frac{1}{\sqrt{1 + 2 a i}})) d a

= \sqrt{π} (\frac{1}{2} (\sqrt{1 - 2 i} + \sqrt{1 + 2 i})) = \sqrt{2 π (1 + \sqrt{5})}

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