Question

Evaluate the following integral

Answer 1

\[\int\limits_{0}^{2\pi} (\sqrt{\frac{1}{3\pi}}+ \sqrt{\frac{1}{6\pi}}*e^{ix}) (\sqrt{\frac{1}{3\pi}}+ \sqrt{\frac{1}{6\pi}}*e^{-ix}) dx\]

Answer 2

The integral should come out to be 1 because \[\sqrt{\frac{1}{6\pi}}e^{ix}*\sqrt{\frac{1}{6\pi}}e^{-ix}=\frac{1}{6\pi}\]
and the ingetegral between 0 and 2pi will be 1/3
and if we do the same for
\[\sqrt{\frac{1}{3\pi}}*\sqrt{\frac{1}{3\pi}}=\frac{1}{3\pi}\]
ad the integral between 0 and 2pi for that is 2/3

I don't understand how the other stuff goes to zero when evaluating the integral.

Answer 3

The integrated result is:
\[\frac{x}{3\pi}+\frac{x}{6\pi}+\sqrt{\frac{1}{18\pi ^{2}}}\times \frac{1}{i}e ^{ix}+\sqrt{\frac{1}{18\pi ^{2}}}\times-\frac{1}{i}e ^{-ix}\]
The last two terms cancel the reason being that:
\[e ^{x}=e ^{-x}\]
By substitution the value of the definite integral = 1 unit

Answer 4

Oh I see it now thanks!!

Answer 5

You're welcome :)