Derivatives of J(x)=(1/pi) int[-pi/2, pi/2][cos(xsin(@))] d@

[micke]

$\displaystyle J_0 (x) = \frac{1}{\pi }\int\limits_{ - \pi /2}^{\pi /2} {\cos (x\sin \theta )d\theta }$

Show that
$\displaystyle J_0 '' + \frac{1}{x}J_0 ' + J_0 = 0$

I've done the following (don't know if it's possible to calculate any of these integrals):
$\displaystyle J_0 ' = \frac{1}{\pi }\int\limits_{ - \pi /2}^{\pi /2} { - \sin (x\sin \theta )\sin \theta d\theta }$

$\displaystyle J_0 '' = \frac{1}{\pi }\int\limits_{ - \pi /2}^{\pi /2} { - \cos (x\sin \theta )\sin ^2 \theta d\theta }$

So now what?
Thanks!

 

[royhaas]

Form the sum $\displaystyle J''_0+J_0$

 

[micke]

Form the sum $\displaystyle J''_0+J_0$

$\displaystyle J_0 '' + J_0 = \frac{1}{\pi }\int\limits_{ - \pi /2}^{\pi /2} {(\cos (x\sin \theta ) - \cos (x\sin \theta )\sin ^2 \theta )d\theta }$

How will this help?

 

[PAULK]

Form the sum $\displaystyle J''_0+J_0$

$\displaystyle J_0 '' + J_0 = \frac{1}{\pi }\int\limits_{ - \pi /2}^{\pi /2} {(\cos (x\sin \theta ) - \cos (x\sin \theta )\sin ^2 \theta )d\theta }$

How will this help?

......................................
It works! It does help!

Take your J0': (you found it earlier)

--- some day I'll learn TEX, but not today; bear with me ----

INT - sin (x sin t) sin t dt

[BTW, lose that 1/pi factor -- it's irrelevant to this.]

Believe it or not, you can use IBP on this:

u = sin (x sin t), du = x cos t cos(x sin t)

dv = sin t dt, v = - cos t

First term:

uv = - cos t sin(x sin t) [from -pi/2 to pi/2]

= 0

...................
Second term
- INT v du =

+ INT cos t (x cos t cos x sin t)

+ INT cos t (x cos t cos (x sin t))

INT x cos^2(t) cos(x sin t) << that's how you get an x in there.

Now:
1. You divide out the x to get the 1/x J'

2. Replace cos^2(t) with 1 - sin^2(t).

3. You have the J0'' + J0 mentioned earlier.

[I might have blown an minus somewhere, but you'll find it.

 

[micke]

Many thanks! (I wonder how I ever would have seen this by myself...)

 

[PAULK]

Many thanks! (I wonder how I ever would have seen this by myself...)

You would have seen it, I'm sure. You would have said:

How can I get an x to be divided out?
If I differentiate cos(x sin t) or sin (x sint ) w.r.t. t, I'll have a factor of x.
Why would I diff that? Oh, it's part of the integrand, so I'll try IBP.