Evaluating Definite Integrals: A Comparison and Explanation

[yungman]

I want to proof
\int_{-\pi}^0 e^{jz\cos \theta}\cos(m\theta) d\theta=\int_0^{\pi} e^{jz\cos \theta}\cos(m\theta) d\theta
(1) Let $\theta=-\theta\;\Rightarrow\; d\theta=-d\theta,\;\cos(-\theta)=\cos\theta,\;\cos[m(-\theta)]=\cos(m\theta)$
\int_{-\pi}^0 e^{jz\cos \theta}\cos(m\theta) d\theta=-\int_{\pi}^0 e^{jz\cos \theta}\cos(m\theta) d\theta=\int_0^{\pi}e^{jz\cos \theta}\cos(m\theta) d\theta

BUT

(2) Let $\theta=\pi+\theta\;\Rightarrow\; d\theta=d\theta,\;\cos(\pi+\theta)=-\cos\theta$
\int_{-\pi}^0 e^{jz\cos \theta}\cos(m\theta) d\theta=\int_0^{\pi} e^{-jz\cos \theta}\cos(m\pi+m\theta) d\theta
\cos(m\pi+m\theta)=\cos(m\pi)\cos(m\theta)-\sin(m\pi)\sin(m\theta)=(-1)^m \cos(m\theta)
\Rightarrow\;\int_{-\pi}^0 e^{jz\cos \theta}\cos(m\theta) d\theta=\int_0^{\pi} e^{-jz\cos \theta}(-1)^m \cos(m\theta) d\theta

Obviously the two ways give different result. What did I do wrong?

Thanks

 

[vanhees71]

Both of your calculations are correct. Where do you see a contradiction?

 

[yungman]

Both of your calculations are correct. Where do you see a contradiction?

But
\int_0^{\pi}e^{jz\cos \theta}\cos(m\theta) d\theta\neq \int_0^{\pi} e^{-jz\cos \theta}(-1)^m \cos(m\theta) d\theta

Thanks for the reply.

 

[yungman]

Anyone please, I don't think I did anything wrong.

 

[Mandelbroth]

But
\int_0^{\pi}e^{jz\cos \theta}\cos(m\theta) d\theta\neq \int_0^{\pi} e^{-jz\cos \theta}(-1)^m \cos(m\theta) d\theta

Thanks for the reply.

Could you give a counterexample for me? I'm feeling tired and lazy right now, but both your derivations, at a glance, look okay to me.

 

[yungman]

Could you give a counterexample for me? I'm feeling tired and lazy right now, but both your derivations, at a glance, look okay to me.

I don't have counter example, I am confused enough already.


This identity is definitely true given but many books:
J_m(z)=\frac{j^{-m}}{2\pi}\int_{-\pi}^{\pi}e^{jz\cos u}\cos(m\theta)d\theta=\frac{j^{-m}}{\pi}\int_0^{\pi} e^{jz\cos \theta}\cos(m\theta) d\theta
J_m(z)=\frac{j^{-m}}{2\pi}\int_{-\pi}^{\pi}e^{jz\cos u}\cos(m\theta)d\theta=\frac{j^{-m}}{2\pi}\int_0^{\pi} e^{jz\cos \theta}\cos(m\theta) d\theta+\frac{j^{-m}}{2\pi}\int_{-\pi}^{0} e^{jz\cos \theta}\cos(m\theta) d\theta

The original question is regarding the second integral.

Thanks

 

[Mandelbroth]

I don't have counter example, I am confused enough already.

Can you give an $m$ such that the two integrals are not equal? Checking the trivial case $m=0$, we get that the two integrals are equal. You say that the two ways "obviously" give a different result, but I'm pretty sure the two integrals are equal. I want to know why you think they wouldn't be.

 

[yungman]

Can you give an $m$ such that the two integrals are not equal? Checking the trivial case $m=0$, we get that the two integrals are equal. You say that the two ways "obviously" give a different result, but I'm pretty sure the two integrals are equal. I want to know why you think they wouldn't be.

But for m=1

\int_0^{\pi} e^{-jz\cos \theta}(-1) \cos(\theta) d\theta=-\int_0^{\pi} \frac{1}{e^{jz\cos \theta}} \cos(\theta) d\theta

\int_0^{\pi}e^{jz\cos \theta}\cos(\theta) d\theta\neq -\int_0^{\pi} \frac{1}{e^{jz\cos \theta}} \cos(\theta) d\theta

for m=3

\int_0^{\pi} e^{-jz\cos \theta}(-1) \cos(3\theta) d\theta=-\int_0^{\pi} \frac{1}{e^{jz\cos \theta}} \cos(3\theta) d\theta

\int_0^{\pi}e^{jz\cos \theta}\cos(3\theta) d\theta\neq -\int_0^{\pi} \frac{1}{e^{jz\cos \theta}} \cos(3\theta) d\theta

 

[Mandelbroth]

But for m=1
[...]
\int_0^{\pi}e^{jz\cos \theta}\cos(\theta) d\theta\neq -\int_0^{\pi} \frac{1}{e^{jz\cos \theta}} \cos(\theta) d\theta

This is not true. For $m=1$, they are both equal to $j\pi J_1(z)$, where $J_1$ is a Bessel function of the first kind. The integrals are also equal for $m=3$.

 

[yungman]

This is not true. For $m=1$, they are equal to $j\pi J_1(z)$, where $J_1$ is a Bessel function of the first kind. The integrals are also equal for $m=3$.

Please explain a little more, I don't get what you mean. All I am looking is that part of the equation is not equal and I don't see a reason.

The important thing is not the m. It's the $\;e^{jx\cos\theta}\;$ can never equal to $\;e^{-jx\cos\theta}\;$ except $\;\theta=\pi/2$

 

[Mandelbroth]

Please explain a little more, I don't get what you mean. All I am looking is that part of the equation is not equal and I don't see a reason.

The important thing is not the m. It's the $\;e^{jx\cos\theta}\;$ can never equal to $\;e^{-jx\cos\theta}\;$ except $\;\theta=\pi/2$

You're just looking at the integrands. You CAN'T do that. Different integrands can have the same definite integrals.

 

[micromass]

Let's take a look at the graphs. Here is the graph for the real part for $z=3$ and $m=3$:

As you see, the graphs won't be equal, but the integrals will be. I hope this reassures you that your result is correct.

 

[yungman]

You're just looking at the integrands. You CAN'T do that. Different integrands can have the same definite integrals.

Sorry I still don't get this.
It is given:
J_m(z)=\frac{j^{-m}}{2\pi}\int_{-\pi}^{\pi}e^{jz\cos u}\cos(m\theta)d\theta=\frac{j^{-m}}{\pi}\int_0^{\pi} e^{jz\cos \theta}\cos(m\theta) d\theta

From that we have:
J_m(z)=\frac{j^{-m}}{2\pi}\int_{-\pi}^{\pi}e^{jz\cos u}\cos(m\theta)d\theta=\frac{j^{-m}}{2\pi}\int_0^{\pi} e^{jz\cos \theta}\cos(m\theta) d\theta+\frac{j^{-m}}{2\pi}\int_{-\pi}^{0} e^{jz\cos \theta}\cos(m\theta) d\theta

This mean it has to be:
\frac{j^{-m}}{2\pi}\int_0^{\pi} e^{jz\cos \theta}\cos(m\theta) d\theta=\frac{j^{-m}}{2\pi}\int_{-\pi}^{0} e^{jz\cos \theta}\cos(m\theta) d\theta

From my original post I use two different change of variables and get two different result, and:
\frac{j^{-m}}{2\pi}\int_0^{\pi}e^{jz\cos \theta}\cos(m\theta) d\theta\neq \frac{j^{-m}}{2\pi}\int_0^{\pi} \frac{1}{e^{jz\cos \theta}}(-1)^m \cos(m\theta) d\theta

Or
J_m(z)=\frac{j^{-m}}{\pi}\int_0^{\pi} e^{jz\cos \theta}\cos(m\theta) d\theta\neq\frac{j^{-m}}{2\pi}\int_0^{\pi} e^{jz\cos \theta}\cos(m\theta) d\theta+\frac{j^{-m}}{2\pi}\int_{0}^{\pi} \frac{1}{e^{jz\cos \theta}}(-1)^m\cos(m\theta) d\theta

This is regardless of whether it's a Bessel function or not.

I am confused.

Thanks

 

[micromass]

From my original post I use two different change of variables and get two different result, and:
\frac{j^{-m}}{2\pi}\int_0^{\pi}e^{jz\cos \theta}\cos(m\theta) d\theta\neq \frac{j^{-m}}{2\pi}\int_0^{\pi} \frac{1}{e^{-jz\cos \theta}}(-1)^m \cos(m\theta) d\theta

Why do you think the integrals are not equal? Doesn't my graph show you that they were equal (for a special case, ok)?

Wait, didn't you mean:
\frac{j^{-m}}{2\pi}\int_0^{\pi}e^{jz\cos \theta}\cos(m\theta) d\theta\neq \frac{j^{-m}}{2\pi}\int_0^{\pi} e^{-jz\cos \theta}(-1)^m \cos(m\theta) d\theta
As in your orginal post?

 

[yungman]

Why do you think the integrals are not equal? Doesn't my graph show you that they were equal (for a special case, ok)?

Wait, didn't you mean:
\frac{j^{-m}}{2\pi}\int_0^{\pi}e^{jz\cos \theta}\cos(m\theta) d\theta\neq \frac{j^{-m}}{2\pi}\int_0^{\pi} e^{-jz\cos \theta}(-1)^m \cos(m\theta) d\theta
As in your orginal post?

Yes, I am trying to derive the equation:
J_m(z)=\frac{j^{-m}}{2\pi}\int_{-\pi}^{\pi}e^{jz\cos u}\cos(m\theta)d\theta=\frac{j^{-m}}{\pi}\int_0^{\pi} e^{jz\cos \theta}\cos(m\theta) d\theta
AND I can't get there if I cannot prove
\frac{j^{-m}}{2\pi}\int_0^{\pi}e^{jz\cos \theta}\cos(m\theta) d\theta\neq \frac{j^{-m}}{2\pi}\int_0^{\pi} e^{-jz\cos \theta}(-1)^m \cos(m\theta) d\thetaMy original post is not even a Bessel function question. It's a change of variable problem...that I use two different change of variable and get two different answer that I cannot prove they are equal.

Thanks for your help.

 

[micromass]

Yes, I am trying to derive the equation:
J_m(z)=\frac{j^{-m}}{2\pi}\int_{-\pi}^{\pi}e^{jz\cos u}\cos(m\theta)d\theta=\frac{j^{-m}}{\pi}\int_0^{\pi} e^{jz\cos \theta}\cos(m\theta) d\theta
AND I can't get there if I cannot prove
\frac{j^{-m}}{2\pi}\int_0^{\pi}e^{jz\cos \theta}\cos(m\theta) d\theta\neq \frac{j^{-m}}{2\pi}\int_0^{\pi} e^{-jz\cos \theta}(-1)^m \cos(m\theta) d\theta

Bottom line, even maybe the result is same, but how do I prove the formula?
Thanks for your help.

I think your post in your OP actually provides a good proof of the formula.

 

[yungman]

I think your post in your OP actually provides a good proof of the formula.

Can you explain more? I am more confused now. How come I have two ways and get two different result and I cannot prove it's the same. This is not even issue with Bessel function. Just a change of variable problem.

Thanks for your help.

 

[micromass]

Can you explain more?

Well, you started from one integral. You did two different substitutions (namely $\theta\rightarrow - \theta$ and $\theta \rightarrow \pi + \theta$) and you obtained two different integrals equal to the first integral. So these two integrals must equal.

 

[yungman]

Well, you started from one integral. You did two different substitutions (namely $\theta\rightarrow - \theta$ and $\theta \rightarrow \pi + \theta$) and you obtained two different integrals equal to the first integral. So these two integrals must equal.

That's how it work? Isn't it pushing it? How do you justify the two equations in my OP are equal? Even you use numerical method, it's not going to be equal. I thought math has to have reason and valid justification. What I showed is very basic change of variables only!

Forget that it's part of the Bessel function. How do you justify to beginning math students that this basic change of variables give you the same solution...as it absolutely does not. Unless you are going to amend that there are exception to the change of variables when comes to some difficult functions.

No matter how you cut it
From my original post I use two different change of variables and get two different result, and:
\int_0^{\pi}e^{jz\cos \theta}\cos(m\theta) d\theta\neq\int_0^{\pi} \frac{1}{e^{jz\cos \theta}}(-1)^m \cos(m\theta) d\theta
As you cannot make:
e^{jz\cos \theta}\neq \frac{1}{e^{jz\cos \theta}}
For all z and $\theta$.

I don't mean to challenge you or others, it just does not make sense.

Thanks

 

[Mandelbroth]

That's how it work? Isn't it pushing it? How do you justify the two equations in my OP are equal? Even you use numerical method, it's not going to be equal. I thought math has to have reason and valid justification. What I showed is very basic change of variables only!

Forget that it's part of the Bessel function. How do you justify to beginning math students that this basic change of variables give you the same solution...as it absolutely does not. Unless you are going to amend that there are exception to the change of variables when comes to some difficult functions.

No matter how you cut it
From my original post I use two different change of variables and get two different result, and:
\int_0^{\pi}e^{jz\cos \theta}\cos(m\theta) d\theta\neq\int_0^{\pi} \frac{1}{e^{jz\cos \theta}}(-1)^m \cos(m\theta) d\theta
As you cannot make:
e^{jz\cos \theta}\neq \frac{1}{e^{jz\cos \theta}}
For all z and $\theta$.

I don't mean to challenge you or others, it just does not make sense.

Thanks

So, by your logic, $\displaystyle \int\limits_{[0,2\pi]}\cos{x}\, dx = \int\limits_{[0,2\pi]}\sin{x}\, dx$ is not true because $\sin{x}\neq\cos{x}$?

 

[yungman]

So, by your logic, $\displaystyle \int\limits_{[0,2\pi]}\cos{x}\, dx = \int\limits_{[0,2\pi]}\sin{x}\, dx$ is not true because $\sin{x}\neq\cos{x}$?

I don't know, I am not a math expert by any stretch. I don't mean I don't trust you. Is there a mathematical proof they are equal like all the equations, theorem in textbooks?

In your example, both result in zero. But in my case, the result is not zero.

It's like saying \frac 0 a=\frac 0 b does not mean a=b

But how about \frac k a=\frac k b where k is a non zero value?

I don't know. I just feel there is a more convincing way to proof this.

If
\frac{j^{-m}}{\pi}\int_0^{\pi} e^{jz\cos \theta}\cos(m\theta) d\theta=\frac{j^{-m}}{2\pi}\int_0^{\pi} e^{jz\cos \theta}\cos(m\theta) d\theta+\frac{j^{-m}}{2\pi}\int_{0}^{\pi} \frac{1}{e^{jz\cos \theta}}(-1)^m\cos(m\theta) d\theta

AND if this is a LINEAR equation and it's not zero, is there something about
\int_0^{\pi} e^{jz\cos \theta}\cos(m\theta) d\theta=\int_{0}^{\pi} \frac{1}{e^{jz\cos \theta}}(-1)^m\cos(m\theta) d\theta
For all $\theta$ and z?Thanks

 

[yungman]

This is getting too confusing. I understand the two equation can look very different but the result is the same for all value. This is like

e^x=\sum_0^{\infty}\frac {x^k}{k!}
Even the two side is totally different, but there is a reason, because it is a power series relation and there is a definite way to go from the left side to the right side. So if the assertion is true in my OP, is there a step by step proof for that? there has to be a logical step by step derivation in math...at least at this more basic level of change of variables.

Is there any way than proof by plotting a graph?

 

[yungman]

Anyone please?

 

[dirk_mec1]

What is it that you do not understand? You used a change of variable and both of your results MUST be the same since you have not performed any illegal operation.

 

[yungman]

What is it that you do not understand? You used a change of variable and both of your results MUST be the same since you have not performed any illegal operation.

Then how come I cannot transform from one result to the other?

 

[Boorglar]

I think you're mistaking equality of integrals with equality of integrands.

Just because \int_{a}^{b}f(x)dx = \int_{a}^{b}g(x)dx does not mean that f(x) = g(x) for all x in [a,b].

Example:

\int_{0}^{1} 2xdx = 1 = \int_{0}^{1} 3x^2dx

But clearly 2x ≠ 3x^2 on [0,1].

The point is that the definite integral over an interval is not a local property. Two entirely different functions may happen to have the same integral over a given interval.

 

[Mandelbroth]

Then how come I cannot transform from one result to the other?

You don't necessarily have to be able to "transform" one integrand into the other. They don't have to be equal. For example, consider that $\displaystyle \int\limits_{[0,\pi]}\frac{2}{\pi}~dx=\int\limits_{[0,\pi]}\sin{x}~dx$. Definite integrals don't necessarily have to have the same integrand to be equal.

 

[yungman]

Thanks for all the replies.

 

[Mandelbroth]

Thanks for all the replies.

You're welcome. We're happy to help.

 

[yungman]

I finally got the answer from Math Forum by Mathman

v=\pi-u\;\Rightarrow \;u=\pi-v,\;du=-dv

\int_{0}^{\pi}e^{jx\cos (u)}\cos(mu)du=-\int_{\pi}^{0}e^{jx\cos (\pi-v)}\cos(m\pi-mv)dv=\int_{0}^{\pi}e^{-jx\cos (v)}(-1)^m\cos(mv)dv

\theta=v\;\Rightarrow \;\int_{0}^{\pi}e^{-jx\cos (v)}(-1)^m\cos(mv)dv=\int_{0}^{\pi}e^{-jx\cos (\theta)}(-1)^m\cos(m\theta)d\theta

This is what I am looking for, a step by step proof that the two change of variable give the same exact answer. I know there got to be a way to proof identity.