Question

I'd be very grateful if someone could start me off on this:
Express
exp( (2x) / π ) cos (3x)
as the real part of a single complex exponential. Hence find
∫ exp( (2x) / π ) cos 3x dx . (the limits I can't draw but are between 0 and π) @Physics

Answer 1

The formula is this: http://www.wolframalpha.com/input/?i=exp%28+%282x%29+%2F+%CF%80+%29+cos+%283x%29#
The integral: http://www.wolframalpha.com/input/?i=integrate++between+0+and+pi+%28exp%28+%282x%29+%2F+%CF%80+%29+cos+%283x%29%29#

Answer 2

I think it wants cos (3x) rewriting as an exponential.

Answer 3

\[1/2 (e^{-3 i x}+ e^{3 i x})\]

Answer 4

yes, then that turns out to be \[1/2(2e^{2x/\pi}+e^{-3 i x}+e^{3 i x })\] which is obviously equal to\[\cos(3x) + \sinh (2x/ \pi) + \cosh (2x/\pi)\]Which would be dead easy to integrate. Do you think that's right?

Answer 5

So that gives as the final answer\[(\pi ^2 -4 e^2)/(2 \pi)\]

Answer 6

integrating gave \[\left[ -3x \sin (3x) + 2/\pi \cosh (2x/ \pi) +2/\pi \sinh(2x/\pi) \right]_{0}^{\pi}\]

Answer 7

What being asked here is this: what is the exponent f(x) such that
Re [ exp(f(x)) ] = exp( (2x) / π ) cos (3x)

And is clearly f(x) = ( 2/ π + 3i) x

Answer 8

Hence the integral of exp( (2x) / π ) cos (3x) is the integral of the real part of
exp( ( 2/ π + 3i) x )
which will turn out to be the real part of the integral of the same.
I.e.,
\[ Re \left[ \int_0^{\pi} e^{(2/\pi + 3i)x} dx \right] \]

Answer 9

let cos3x be = coskx

coskx= (1/2i).[(exp(ikx)+exp(-ikx)] put this into the integral

S exp(2x/pi). (1/2i).[(exp(ikx)+exp(-ikx)].dx (1/2i) is not the variable so goes out.
(1/2i) S exp(2x/pi).[(exp(ikx)+exp(-ikx)].dx
(1/2i) S [exp(2x/pi).(exp(ikx)+exp(2x/pi)exp(-ikx)].dx

after thıs its quite easy
just integral of an expo

Answer 10

The question asks: "Express exp( (2x) / π ) cos (3x) as the real part of a single complex exponential."

That's what I've done.

Answer 11

I still don't get it. Which bit is the real part of the complex exponential and which part is the complex exponential?

Answer 12

exp( (2x) / π ) cos (3x) = Re [ exp( (2x) / π ) . exp(3xi) ]
= Re [ exp( (2/π+3i)x ) ]

Answer 13

Oh right. I get it now. so you're saying exp( (2x) / π ) cos (3x) is only some of the truth. There's a complex bit which we've left out. The full thing is: exp( (2x) / π ) . exp(3xi) (the complex exponential) which can be integrated easily. Thank you so much.

Answer 14

yes.

Answer 15

If I could give you Graham's number of medals, I would.

Answer 16

One will suffice ;-)