Mathematical Yin-Yang

Calculus Level 4

If 1 x = 2 π ∫ 0 ∞ e − a 2 x d α \displaystyle \frac { 1 }{ \sqrt { x } } =\frac { 2 }{ \sqrt { \pi } } \int _{ 0 }^{ \infty }{ e^{ -a^2x }d\alpha } for α > 0 \alpha>0 , evaluate ∫ 0 ∞ cos ⁡ x d x x − ∫ 0 ∞ sin ⁡ x d x x \large \int _{ 0 }^{ \infty }{ \frac {\cos{x} \ dx }{\sqrt{x} } } -\int _{ 0 }^{ \infty }{ \frac{\sin{x} \ dx}{\sqrt{x}} }


The answer is 0.

John M. by John M. , 24, USA

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1 solution

First Last
Jun 5, 2017

Using a general Gaussian ∫ 0 ∞ x b e − a x n d x = Γ ( 1 + b n ) a 1 + b n n (sub x = a x n and use gamma function definition) \displaystyle\int_0^\infty x^be^{-ax^n}dx=\frac{\Gamma(\frac{1+b}{n})}{a^{\frac{1+b}{n}}n}\quad\text{(sub }x=ax^n\text{ and use gamma function definition)}

I = ∫ 0 ∞ cos ⁡ x x − sin ⁡ x x d x = ∫ 0 ∞ x − 1 2 e i x + e − i x 2 − x − 1 2 e i x − e − i x 2 i d x = Γ ( 1 2 ) ( 1 − i + 1 i 2 − 1 − i − 1 i 2 i ) = \displaystyle I = \int_0^\infty \frac{\cos{x}}{\sqrt{x}}-\frac{\sin{x}}{\sqrt{x}}dx = \int_0^\infty x^\frac{-1}{2}\frac{e^{ix}+e^{-ix}}{2}-x^\frac{-1}{2}\frac{e^{ix}-e^{-ix}}{2i}dx=\Gamma(\frac1{2})(\frac{\frac1{\sqrt{-i}}+\frac1{\sqrt{i}}}{2}-\frac{\frac1{\sqrt{-i}}-\frac1{\sqrt{i}}}{2i})=

Γ ( 1 2 ) ( 1 2 − 1 2 ) = 0 \displaystyle\Gamma(\frac1{2})(\frac1{\sqrt{2}}-\frac1{\sqrt{2}}) = \boxed{0}

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