Easy integral?

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
Compute \(\displaystyle \int_0^{2\pi}e^{\cos(x)}\cos\left(x+\sin(x)\right)dx\)
 
Oct 2009
4,261
1,836
Is the integrand the derivative of

\(\displaystyle e^{\cos(x)} \sin( \sin(x) ) \) ?

\(\displaystyle (e^{\cos(x)} \sin( \sin(x) ))'=-\sin x \,e^{\cos x}\sin(\sin x)+e^{\cos x}\cos(\sin x) \cos x \) \(\displaystyle =e^{\cos x}\left(\cos(\sin x)\cos x-\sin x\,\sin(\sin x)\right)=\)... \(\displaystyle e^{\cos x}\cos(x+\sin x)\) !! Oh, dear hollie mollie: yes, it is!

A question: how did you come up with it? Some program or some insight?

Tonio
 

simplependulum

MHF Hall of Honor
Jan 2009
715
427
I write \(\displaystyle e^{\cos(x)} \cos( x + \sin(x) )\)

\(\displaystyle = Re[ e^{\cos(x)}e^{ix} e^{i\sin(x)} ] \)

\(\displaystyle = Re[ e^{e^{ix}} e^{ix} ] \)

then substitute \(\displaystyle u = e^{ix} \)

At first i did it by integration by parts but the steps are quite long so i was not sure at that time ... Now , i am sure it is true with your help .(Happy)
 
  • Like
Reactions: Bruno J.

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
\(\displaystyle (e^{\cos(x)} \sin( \sin(x) ))'=-\sin x \,e^{\cos x}\sin(\sin x)+e^{\cos x}\cos(\sin x) \cos x \) \(\displaystyle =e^{\cos x}\left(\cos(\sin x)\cos x-\sin x\,\sin(\sin x)\right)=\)... \(\displaystyle e^{\cos x}\cos(x+\sin x)\) !! Oh, dear hollie mollie: yes, it is!

A question: how did you come up with it? Some program or some insight?

Tonio
Just tinkering

I write \(\displaystyle e^{\cos(x)} \cos( x + \sin(x) )\)

\(\displaystyle = Re[ e^{\cos(x)}e^{ix} e^{i\sin(x)} ] \)

\(\displaystyle = Re[ e^{e^{ix}} e^{ix} ] \)

then substitute \(\displaystyle u = e^{ix} \)

At first i did it by integration by parts but the steps are quite long so i was not sure at that time ... Now , i am sure it is true with your help .(Happy)
There is another method that involves complex analysis, and it is the way I was hoping someone would do this.


We know that \(\displaystyle \oint_{|z|=1}e^zdz=0\) and thus \(\displaystyle \int_0^{2\pi}e^{e^{it}}\left(e^{it}\right)'dt=\int_0^{2\pi}e^{\cos(t)}\left(\cos(\sin(t))+i\sin(\sin(t))\right)\left(-\sin(t)+i\cos(t)\right)=0\), in particular the real an imaginary parts of that integral must equal zero. Do a little work with trig identities and you will get our integral is the imaginary part of it.
 
May 2009
959
362
let \(\displaystyle I(a) = \int^{2 \pi}_{0} e^{a \cos x} \cos (x + a \sin x) \ dx = Re \Big(\int^{2 \pi}_{0} e^{ae^{ix}}e^{ix} \ dx \Big)\)

then \(\displaystyle I'(a) = Re \Big(\int^{2 \pi}_{0} e^{2ix}e^{ae^{ix}} \ dx\Big) \)

let \(\displaystyle u = ae^{ix} \)

EDIT: then \(\displaystyle I'(a) = Re \Big( -\frac{i}{a^{2}} \int^{a}_{a} ue^{u} \ du \Big) = Re(0) = 0 \)

so \(\displaystyle I(a) = C \)

but \(\displaystyle I(0) = 0\)

which means \(\displaystyle C=0\)

so \(\displaystyle I(a) = 0 \)

and \(\displaystyle \int^{2 \pi}_{0} e^{\cos x} \cos (x + \sin x) \ dx = I(1) = 0 \)
 
Last edited: