How can I evaluate this non-elementary integral?

[yungman]

Evaluate
\int_0^{\pi}\cos(a\sin\theta)d\theta

This is my work so far and I don't know how to proceed or is this even correct

Let $u=a\sin\theta \Rightarrow\; \frac {du}{a\cos\theta}=d\theta$
\int_0^{\pi}\cos(a\sin\theta)d\theta=\int\frac{\cos u}{a\cos\theta}du
But I cannot set the limit as both are zero! So I just skip this step and put the interval back in after calculation.

Let $W=a\cos\theta\Rightarrow\;dW=-a\sin\theta$. Let $dv=\frac {1}{\cos\theta}=\sec\theta\Rightarrow\;v=\int\sec\theta d\theta=ln|\sec \theta+\tan\theta|$

This very complicated to substitute.


So I tried this
\int\frac{\cos u}{a\cos\theta}du=\frac {1}{a} \int \frac{\cos u}{\sqrt{1-\sin^2\theta}}du= \frac {1}{x} \int \frac{\cos u}{\sqrt{1-(\frac{u}{a})^2}}du
Still it is very hard. Is that any better way?

Thanks

 

[SteamKing]

Your chosen substitution seems a bit sketchy. Instead of getting rid of x or θ, you have added a third variable u to the mix. I don't see where you are simplifying the original integral or getting it into a recognized form.

 

[yungman]

Your chosen substitution seems a bit sketchy. Instead of getting rid of x or θ, you have added a third variable u to the mix. I don't see where you are simplifying the original integral or getting it into a recognized form.

Thanks


I tried Letting $u=\sin\theta\Rightarrow\; du=-\cos\theta d\theta$.

\int_0^{\pi}\cos(a\sin\theta)d\theta=-\int\frac{\cos au}{\cos\theta}du=-\int\frac{\cos xu}{\sqrt{1-u^2}}du=-\frac {1}{\sqrt {2}}\left[\int\frac{\cos au}{\sqrt{1+u}}du+\int\frac{\cos au}{\sqrt{1-u}}du\right]
But I cannot set the limit as both are zero! I tried going through and it is not simpler either.

 

[LCKurtz]

This is not an elementary integral. Maple gives an answer $\pi$BesselJ(0,x). You can read about properties of the Bessel functions at http://www.efunda.com/math/bessel/bessel.cfm where you will see similar results near the bottom. There are many other places on the internet where you can read about Bessel functions.

 

[yungman]

This is not an elementary integral. Maple gives an answer $\pi$BesselJ(0,x). You can read about properties of the Bessel functions at http://www.efunda.com/math/bessel/bessel.cfm where you will see similar results near the bottom. There are many other places on the internet where you can read about Bessel functions.

Yes, this is part of the integral representation of Bessel Function. So looking up or using online calculator is not my first choice yet as it defeats the integral representation of Bessel Function.

I just look up my notes in the pass example and simplified one more step. I am hoping I can solve this.

Thanks

 

[Goa'uld]

I don't think it's possible to get a result of finite elementary functions involving x. Plus, if both the limits of the definite integral are equal after a substitution, isn't the integral zero?

 

[yungman]

I don't think it's possible to get a result of finite elementary functions involving x. Plus, if both the limits of the definite integral are equal after a substitution, isn't the integral zero?

Actually x is treated as a constant here. $\theta$ is the only variable. I went back and change all the "x" to "a" to avoid confusion. x is only being used after deriving the integral representation of the Bessel Function of x.

Thanks

 

[Goa'uld]

I know. You said you're hoping you can solve this, but I don't know what you're trying to solve.

 

[yungman]

I know. You said you're hoping you can solve this, but I don't know what you're trying to solve.

The final equation is
J_m=\frac {1}{\pi}\int_0^{\pi} \cos(x\sin\theta-m\theta)d\theta \;\hbox { for } \;m=,0,1,2,3...

So with that, if given x and m, you find $J_m(x)$.

So if you want to find $J_m(x)$, you need to solve the equation, not by looking up or use online calculator.

I since simplify further:

\int_0^{\pi}\cos(a\sin\theta)d\theta=-\int\frac{\cos au}{\cos\theta}du=-\int\frac{\cos xu}{\sqrt{1-u^2}}du=-\frac {1}{\sqrt {2}}\left[\int\frac{\cos au}{\sqrt{1+u}}du+\int\frac{\cos au}{\sqrt{1-u}}du\right]

Now we simplify $\int\frac{\cos au}{\sqrt{1+u}}du$

Let $v=a(1+u)\;\Rightarrow\; u=\frac{v}{a}-1,\;dv=adu$

\int\frac{\cos au}{\sqrt{1+u}}du=\int\frac{\cos (v-a)}{\sqrt{\frac {v}{a}}}du=\frac{1}{a\sqrt{a}}\int \frac {\cos(v-a)}{\sqrt{v}}dv=\frac{1}{a\sqrt{a}} \int \frac{\cos v\cos a}{\sqrt{v}}dv+\frac{1}{a\sqrt{a}} \int \frac{\sin v\sin a} {\sqrt{v}} dv

\Rightarrow\;\int\frac{\cos au}{\sqrt{1+u}}du=\frac{\cos a}{a\sqrt{a}} \int \frac{\cos v}{\sqrt{v}}dv+\frac{\sin a}{a\sqrt{a}} \int \frac{\sin v} {\sqrt{v}} dv

The other half can be simplified simiarly. Anyway to solve this? I tried integral by parts, won't work.

 

[skiller]

I tried Letting $u=\sin\theta\Rightarrow\; du=-\cos\theta d\theta$.

Where does that minus sign come from?

 

[Infrared]

I think i have solved it a different way- using series. I just need a quick calculus fact:
\int sin^n(x) dx=-\frac{1}{n}sin^{n-1}(x)cos(x)+\frac{n-1}{n} \int sin^{n-2}(x) dx. This can be proven using integration by parts. Replacing n with 2n and putting in bounds of 0 and π gives the formula \int_0^\pi sin^{2n}(x) dx= \frac{2n-1}{2n} \int_0^\pi sin^{2n-2}(x) dx.

OK, \int_0^\pi cos(x sin(\theta)) d\theta= \int_0^\pi 1-\frac{x^2 sin^2(\theta)}{2!}+\frac{x^4 sin^4(\theta)}{4!}-\frac{x^6 sin^6(\theta)}{6!}... d\theta \\ = \int_0^\pi 1 d\theta - \frac{x^2}{2!}\int_0^\pi sin^2(\theta) d\theta+ \frac{x^4}{4!}\int_0^\pi sin^4(\theta) d\theta-\frac{x^6}{6!}\int_0^\pi sin^6(\theta) d\theta...
(If I recall correctly from my calc class, the integral of a convergent power series can be broken up like this)
Ok, so by using the formula at the top of my post \int_0^\pi sin^2(\theta)d\theta=\pi /2 , \int_0^\pi sin^4(\theta)d\theta = \frac{3}{4} \cdot \frac{\pi}{2}=\frac{3\pi}{8} and in general \int_0^\pi sin^{2n}(\theta) d\theta= \frac{\pi (1)(3)(5)...(2n-1)}{(2)(4)(6)...(2n)}

We want to evaluate \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \int_0^\pi sin^{2n}(\theta) d\theta=\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \frac{\pi (1)(3)(5)...(2n-1)}{(2)(4)(6)...(2n)}= \\ \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n} \pi (1)(3)(5)...(2n-1)}{(2)(4)(6)...(2n)(2n!)}= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n} \pi}{(2^2)(4^2)(6^2)...((2n)^2)}, which you will recognize as \pi J_0(x) by factoring out a π

Edit: I see that you replaced the x with an a in your original post. This should make no difference other than a change of variables.

 

[yungman]

Where does that minus sign come from?

Wasn't think!

 

[yungman]

I think i have solved it a different way- using series. I just need a quick calculus fact:
\int sin^n(x) dx=-\frac{1}{n}sin^{n-1}(x)cos(x)+\frac{n-1}{n} \int sin^{n-2}(x) dx. This can be proven using integration by parts. Replacing n with 2n and putting in bounds of 0 and π gives the formula \int_0^\pi sin^{2n}(x) dx= \frac{2n-1}{2n} \int_0^\pi sin^{2n-2}(x) dx.

OK, \int_0^\pi cos(x sin(\theta)) d\theta= \int_0^\pi 1-\frac{x^2 sin^2(\theta)}{2!}+\frac{x^4 sin^4(\theta)}{4!}-\frac{x^6 sin^6(\theta)}{6!}... d\theta \\ = \int_0^\pi 1 d\theta - \frac{x^2}{2!}\int_0^\pi sin^2(\theta) d\theta+ \frac{x^4}{4!}\int_0^\pi sin^4(\theta) d\theta-\frac{x^6}{6!}\int_0^\pi sin^6(\theta) d\theta...
(If I recall correctly from my calc class, the integral of a convergent power series can be broken up like this)
Ok, so by using the formula at the top of my post \int_0^\pi sin^2(\theta)d\theta=\pi /2 , \int_0^\pi sin^4(\theta)d\theta = \frac{3}{4} \cdot \frac{\pi}{2}=\frac{3\pi}{8} and in general \int_0^\pi sin^{2n}(\theta) d\theta= \frac{\pi (1)(3)(5)...(2n-1)}{(2)(4)(6)...(2n)}=

We want to evaluate \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \int_0^\pi sin^{2n}(\theta) d\theta=\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \frac{\pi (1)(3)(5)...(2n-1)}{(2)(4)(6)...(2n)}= \\ \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n} \pi (1)(3)(5)...(2n-1)}{(2)(4)(6)...(2n)(2n!)}= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n} \pi}{(2^2)(4^2)(6^2)...((2n)^2)}, which you will recognize as \pi J_0(x) by factoring out a π

Edit: I see that you replaced the x with an a in your original post. This should make no difference other than a change of variables.

Thanks, Let me work it out.

Alan

 

[Ray Vickson]

Evaluate
\int_0^{\pi}\cos(a\sin\theta)d\theta

This is my work so far and I don't know how to proceed or is this even correct

Let $u=a\sin\theta \Rightarrow\; \frac {du}{a\cos\theta}=d\theta$
\int_0^{\pi}\cos(a\sin\theta)d\theta=\int\frac{\cos u}{a\cos\theta}du
But I cannot set the limit as both are zero! So I just skip this step and put the interval back in after calculation.

Let $W=a\cos\theta\Rightarrow\;dW=-a\sin\theta$. Let $dv=\frac {1}{\cos\theta}=\sec\theta\Rightarrow\;v=\int\sec\theta d\theta=ln|\sec \theta+\tan\theta|$

This very complicated to substitute.


So I tried this
\int\frac{\cos u}{a\cos\theta}du=\frac {1}{a} \int \frac{\cos u}{\sqrt{1-\sin^2\theta}}du= \frac {1}{x} \int \frac{\cos u}{\sqrt{1-(\frac{u}{a})^2}}du
Still it is very hard. Is that any better way?

Thanks

Your transformation $u = a \sin(\theta)$ is not one-to-one (i.e., not monotone), and that is why you need to be really careful when doing a definite integral. You should write
\int_0^{\pi}\cos(a\sin\theta)d\theta = 2 \int_0^{\pi/2}\cos(a\sin\theta)d\theta, and then do your change of variable in the integral on the right. Your integration limits will be correct now.

 

[yungman]

Your transformation $u = a \sin(\theta)$ is not one-to-one (i.e., not monotone), and that is why you need to be really careful when doing a definite integral. You should write
\int_0^{\pi}\cos(a\sin\theta)d\theta = 2 \int_0^{\pi/2}\cos(a\sin\theta)d\theta, and then do your change of variable in the integral on the right. Your integration limits will be correct now.

Thanks for the response. Can you explain a little? How do you know you can change the upper boundary from $\pi$ to $\frac {\pi}{2}$? Is it because cosine is positive from 0 to $\pi$ and integration of 0 to $\frac {\pi}{2}$ gives the same value as from $\frac {\pi}{2}$ to $\pi$?

Also if you look at post #9, I think I got quite close. I just need to evaluate

\int \frac{\sin v} {\sqrt{v}} dv\;\hbox { and}\; \int \frac{\cos v}{\sqrt{v}}dv

Can you help?

Thanks

 

[Infrared]

Thanks for the response. Can you explain a little? How do you know you can change the upper boundary from $\pi$ to $\frac {\pi}{2}$? Is it because cosine is positive from 0 to $\pi$ and integration of 0 to $\frac {\pi}{2}$ gives the same value as from $\frac {\pi}{2}$ to $\pi$?

Also if you look at post #9, I think I got quite close. I just need to evaluate

\int \frac{\sin v} {\sqrt{v}} dv\;\hbox { and}\; \int \frac{\cos v}{\sqrt{v}}dv

Can you help?

Thanks

Those integrals are worse to evaluate than the one in the OP. Wolfram alpha gives answers in terms of the Fresnel Integrals which do not look fun to work with. http://en.wikipedia.org/wiki/Fresnel_integral

 

[yungman]

Those integrals are worse to evaluate than the one in the OP. Wolfram alpha gives answers in terms of the Fresnel Integrals which do not look fun to work with. http://en.wikipedia.org/wiki/Fresnel_integral

I looked at your link, I didn't see anything similar.

 

[Infrared]

Here is Wolfram Alpha's answer for your sine integral that uses a Fresnel Integral. http://www.wolframalpha.com/input/?i=integrate+sin(x)/sqrt(x)

 

[yungman]

Here is Wolfram Alpha's answer for your sine integral that uses a Fresnel Integral. http://www.wolframalpha.com/input/?i=integrate+sin(x)/sqrt(x)

Thanks

I guess asking for a conventional evaluation of this problem is asking too much! Both ways end up in series expansions. I know this one has a series expansion as I can incorporates $\sqrt v$ into the series expansion of the $\cos v$. Just hoping to have a cleaner way of evaluating this.

Thanks

 

[Infrared]

Thanks

I guess asking for a conventional evaluation of this problem is asking too much! Both ways end up in series expansions. I know this one has a series expansion as I can incorporates $\sqrt v$ into the series expansion of the $\cos v$. Just hoping to have a cleaner way of evaluating this.

Thanks

I didn't find anything cleaner than my post #11 (which isn't as bad as it might look), but keep looking. I doubt that there is anything better than a series expansion since the solution isn't elementary.

 

[yungman]

I didn't find anything cleaner than my post #11 (which isn't as bad as it might look), but keep looking. I doubt that there is anything better than a series expansion since the solution isn't elementary.

Thanks, you help a lot. I really appreciate this.