Integral of Cosine

Calculus Level 3

0 cos ( x ) 2 x d x = ? \large \int_{0}^\infty \dfrac{\cos(x)}{2^x}dx = \, ?

1 1 2 2 1 1 ln 2 \frac{1}{1-\ln2} ln 2 1 + ( ln 2 ) 2 \frac{\ln2}{1+(\ln2)^2} ln ( 2 ) \ln(2) 1 1 + ( ln 2 ) 2 \frac{1}{1+(\ln2)^2} \infty

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1 solution

Chew-Seong Cheong
Jul 13, 2016

I = 0 cos x 2 x d x = 0 cos ( x ) 2 x d x = 0 ( e i x ) 2 x d x = ( 0 2 x e i x d x ) = ( 0 e ( i + ln 2 ) x d x ) = ( e ( i + ln 2 ) x i + ln 2 0 ) = ( 1 i + ln 2 ) = ( ln 2 i 1 + ( ln 2 ) 2 ) = ln 2 1 + ( ln 2 ) 2 \begin{aligned} I & = \int_0^\infty \frac {\cos x}{2^x} dx \\ & = \int_0^\infty \frac {\cos (-x)}{2^x} dx \\ & = \int_0^\infty \frac {\Re \left( e^{-ix}\right)}{2^x} dx \\ & = \Re \left( \int_0^\infty 2^{-x} e^{-ix} \ dx \right) \\ & = \Re \left( \int_0^\infty e^{-(i+\ln 2)x} \ dx \right) \\ & = \Re \left( - \frac {e^{-(i+\ln 2)x}}{i+\ln 2} \bigg|_0^\infty \right) \\ & = \Re \left(\frac 1{i+\ln 2} \right) \\ & = \Re \left(\frac {\ln 2 - i}{1+(\ln 2)^2} \right) \\ & = \boxed{\dfrac {\ln 2}{1+(\ln 2)^2}} \end{aligned}

sorry but what is that 'R' thing?

Eric Chan - 4 years, 10 months ago

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The real part of a complex number, sometimes it is Re but LaTex shows it as \Re . And imaginary Im is shown as \Im .

Chew-Seong Cheong - 4 years, 10 months ago

It's the symbol for the real part of a complex number.

Vikram Sarkar - 8 months, 1 week ago

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