# Homework Help: Question on definite integral

1. Jul 24, 2013

### 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

2. Jul 24, 2013

### vanhees71

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

3. Jul 24, 2013

### yungman

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

4. Jul 24, 2013

### yungman

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

5. Jul 24, 2013

### Mandelbroth

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.

6. Jul 24, 2013

### yungman

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

7. Jul 24, 2013

### Mandelbroth

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.

8. Jul 24, 2013

### yungman

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

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

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

Last edited: Jul 24, 2013
9. Jul 24, 2013

### Mandelbroth

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 $m=3$.

10. Jul 24, 2013

### yungman

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 $\;\theta=\pi/2$

11. Jul 24, 2013

### Mandelbroth

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

12. Jul 24, 2013

### 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.

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13. Jul 24, 2013

### yungman

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

14. Jul 24, 2013

### micromass

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:
$$\frac{j^{-m}}{2\pi}\int_0^{\pi}e^{jz\cos \theta}\cos(m\theta) d\theta\neq \frac{j^{-m}}{2\pi}\int_0^{\pi} e^{-jz\cos \theta}(-1)^m \cos(m\theta) d\theta$$

15. Jul 24, 2013

### yungman

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
$$\frac{j^{-m}}{2\pi}\int_0^{\pi}e^{jz\cos \theta}\cos(m\theta) d\theta\neq \frac{j^{-m}}{2\pi}\int_0^{\pi} e^{-jz\cos \theta}(-1)^m \cos(m\theta) d\theta$$

My 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.

Last edited: Jul 24, 2013
16. Jul 24, 2013

### micromass

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

17. Jul 24, 2013

### yungman

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.

Last edited: Jul 24, 2013
18. Jul 24, 2013

### micromass

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.

19. Jul 24, 2013

### yungman

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

Last edited: Jul 24, 2013
20. Jul 24, 2013

### Mandelbroth

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 $\sin{x}\neq\cos{x}$?

21. Jul 24, 2013

### yungman

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 text books?

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

Last edited: Jul 24, 2013
22. Jul 24, 2013

### 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?

Last edited: Jul 24, 2013
23. Jul 25, 2013

### yungman

24. Jul 25, 2013

### 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.

25. Jul 25, 2013

### yungman

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