How to prove this integral is 0 ?

[qsk]

For this function : Exp[-Cos[x]]*Cos[x/3]], integrated from 0 to 3*Pi. The numerical result suggests that this integral may gives 0. However, I couldn't find a way to analytically show it's 0. Could anyone tell me how to prove this ?

 

[mathman]

If you look at the graph (analytically) you will see that one side the midpoint of the domain (\frac{3\pi}{2}) is the same as the other with a sign switch. Therefore the integrals over each half cancel.

 

[HallsofIvy]

If you make the change of variable u= x- 3\pi/2 (the midpoint of "0 to 3pi") then x= u+ 3\pi/2 so cos(x)= cos(u+ 3\pi/2)= sin(u) and cos(x/3)= cos(u/3+ \pi/2)= -sin(u/3). Of course, dx= du so the integral becomes -\int_{-3\pi/2}^{3\pi/2} e^{sin(u)}sin(u) du. Now the fact that sine is an "odd" function, sin(-x)= -sin(x) should make it clear that the integral is 0.

 

[qsk]

If you look at the graph (analytically) you will see that one side the midpoint of the domain (\frac{3\pi}{2}) is the same as the other with a sign switch. Therefore the integrals over each half cancel.

I know that, but the shape of the two sides looks very different. How can I proof it analytically the integrals of the two sides cancels.

 

[davidmoore63@y]

This doesn't seem easy to prove. It seems that the definite integral of Exp[-Cos[x]]*Cos[x/n]] from 0 to n*Pi is zero for n integer>=2

 

[davidmoore63@y]

Equivalently the definite integral of Exp[-Cos[nx]]*Cos[x] from 0 to pi is zero

 

[davidmoore63@y]

Halls of Ivy, the presence of exp(sin u) makes the conclusion unclear to me. Also I think the integrand should have sin(u/3) not sin u

 

[mathman]

If you look at the graph (analytically) you will see that one side the midpoint of the domain (\frac{3\pi}{2}) is the same as the other with a sign switch. Therefore the integrals over each half cancel.

I took another look and I believe I was wrong. The graphs on the two sides are the same without a sign switch.

 

[willem2]

I took another look and I believe I was wrong. The graphs on the two sides are the same without a sign switch.

The only axis of symmetry of the graph is the line y = 3kπ with k integer.

http://www.wolframalpha.com/input/?i=+Exp[-Cos[x]]*Cos[x/3]&x=0&y=0
Somehow the part above the y-axis with 0 < x < 3π/2 has the same area as the part below the y-axis with 3π/2 < x < 3π, even tough the shapes are completely different.

BTW, mathworld can't solve the integral symbolically and reports that it is equal to 0 with an error of 10^-16

 

[verty]

For $t \ge 0$, let $y(t) = \int_0^{3\pi} t^{- cos(x)} \; cos(x/3) \; dx$.

Testing with Wolfram Alpha suggests $y$ is always zero. Needless to say, this is way beyond my knowledge.

 

[davidmoore63@y]

Sketch proof (apologies no latex):

Integral [0,3pi] exp(-cos(x))*cos(x/3) dx. Put y=x/3
= 3 Integral [0,pi] exp(-cos(3y))*cos(y) dy
Expand exp(cos(3y))= Sum [n=0 to inf] (cos(3y))^n / n!)
Now cos(3y)^n can be expressed as a sum of cosines of multiples of 3y
So we have cos(y) [ a cos 3y + b cos 6y + etc]
which equals a cos y cos 3y + b cosy cos 6y + ...
and using the relationship cos (A-B) + cos(A+B) = 2 cosAcosB
we will then have a simple sum of cosines of multiples of y
Sum (n=1 to inf) of a(n) cos (ny)
The integral of this is Sum (n=1 to inf) of a(n)/n sin (ny)
which when evaluated between 0 and pi gives zero.

This needs to be made rigorous, obviously.

 

[abbas_majidi]

Sketch proof (apologies no latex):

Integral [0,3pi] exp(-cos(x))*cos(x/3) dx. Put y=x/3
= 3 Integral [0,pi] exp(-cos(3y))*cos(y) dy
Expand exp(cos(3y))= Sum [n=0 to inf] (cos(3y))^n / n!)
Now cos(3y)^n can be expressed as a sum of cosines of multiples of 3y
So we have cos(y) [ a cos 3y + b cos 6y + etc]
which equals a cos y cos 3y + b cosy cos 6y + ...
and using the relationship cos (A-B) + cos(A+B) = 2 cosAcosB
we will then have a simple sum of cosines of multiples of y
Sum (n=1 to inf) of a(n) cos (ny)
The integral of this is Sum (n=1 to inf) of a(n)/n sin (ny)
which when evaluated between 0 and pi gives zero.

This needs to be made rigorous, obviously.

Very good. It's a professional answer.