# How to prove this integral is 0 ?

• qsk
In summary, this function seems to give a result of 0, but it's unclear how to prove it. However, if you look at the graph analytically, you will see that the integrals over each half of the domain cancel out, so the result is definitely zero.
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 ?

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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.

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.

mathman said:
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.

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

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

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 said:
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.

jim mcnamara
mathman said:
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

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.

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
davidmoore63@y said:
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.

## 1. How do I prove an integral is 0?

There is no one definitive way to prove an integral is 0, as it depends on the specific integral and the techniques used to solve it. However, a common way to prove an integral is 0 is by using the Fundamental Theorem of Calculus or by using substitution to simplify the integral.

## 2. Can I use integration by parts to prove an integral is 0?

Yes, integration by parts is a common technique used to prove an integral is 0. By choosing the right functions to integrate, one can often simplify the integral and show that it equals 0.

## 3. Is it possible for an integral to be 0 even if the function being integrated is not?

Yes, it is possible for an integral to be 0 even if the function being integrated is not. This can happen if the function has both positive and negative values that cancel each other out when integrated over a certain interval.

## 4. Are there any special cases where proving an integral is 0 is particularly difficult?

Yes, there are certain integrals that can be very challenging to prove are 0. These include integrals involving trigonometric functions, logarithms, and inverse trigonometric functions. In these cases, it may require advanced techniques or knowledge of special properties to successfully prove the integral is 0.

## 5. Can I use a computer program or calculator to prove an integral is 0?

While computer programs and calculators can often help with solving integrals, they should not be relied upon to prove an integral is 0. These tools may provide an approximation or numerical solution, but they cannot provide a formal mathematical proof. It is important to understand and use proper mathematical techniques to prove an integral is 0.

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