Help with one small demonstration

[pocaracas]

Now in LaTeX hope it's ok (the preview sucks):

1. Homework Statement

given the integral

I_n(r,z)= \int_z^r \frac{T_n(\frac{p}{z})T_n(\frac{p}{r})}{p \sqrt{r^2-p^2} \sqrt{p^2-z^2}} dp

where T_n(x) is the chebyshev polynomial of the first kind:
T_n(x) = 2 x T_{n-1}(x) - T_{n-2}(x)
T_0(x)=1
T_1(x)=x

Important detail -> 0 < z < r < 1

prove that:
I_{n+1} = I_{n-1}
and
I_0 = I_1 = \frac{\pi}{2}3. The Attempt at a Solution

I_0 = I_1 = \frac{\pi}{2} is relatively simple. I just changed variables y= p^2 and let Mathematica do the rest.

The other proof if giving me a hard time.

Substituting the T_n(x) = 2 x T_{n-1}(x)- T_{n-2}(x), I got:

I_{n+1}(r,z) = \int_z^r \frac{4p T_n(\frac{p}{z}) T_n(\frac{p}{r})} {\sqrt{r^2-p^2} \sqrt{p^2-z^2}} dp -

- \int_z^r \frac{2r T_n(\frac{p}{z})T_{n-1}(\frac{p}{r})}{\sqrt{r^2-p^2} \sqrt{p^2-z^2}} dp -

- \int_z^r\frac{2z T_{n-1}(\frac{p}{z}) T_n(\frac{p}{r})}{\sqrt{r^2-p^2} \sqrt{p^2-z^2}} dp + I_{n-1}(r,z)

Now, the remainder integrals should be zero, but I can't figure out how to get there.
If it helps, I've found this relation:
xT_n(x) = \frac{1}{2}(T_{n+1}(x) + T_{n-1}(x))

Any help will be very welcome!

 

[Defennder]

This is extremely hard to read. Could you write in Latex?

 

[pocaracas]

This is extremely hard to read. Could you write in Latex?

I could, but I need to know how to add latex code here.

 

[Pomico]

There is a \sum symbol in the tool bar above the reply box. Click on this and a latex reference should appear :)

 

[pocaracas]

I just noticed that T_n(x) = 2 x T_{n-1}(x) - T_{n-2}(x) is just the same as
xT_n(x) = \frac{1}{2}(T_{n+1}(x) + T_{n-1}(x))

:(

 

[pocaracas]

There is a \sum symbol in the tool bar above the reply box. Click on this and a latex reference should appear :)

thanks man!

 

[pocaracas]

no ideas? :(