I was messing around with the [tex]\theta[/tex] equation of hydrogen atom. OK, the equation is a Legendre differential equation, which has solutions of Legendre polynomials. I haven't studied them before, so I decided to take closed look and began working on the most simple type of Legendre DE. And the story is as follows:

The book says:

[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = \delta_{mn} \frac{2}{2n+1}[/tex]

So I sat and tried derieving it. First, I gather an inventory that might be useful:

[tex](1-x^2)P_n''(x) - 2xP_n'(x) + n(n+1) = 0[/tex]

[tex][(1-x^2)P_n'(x)]' = -n(n+1)P_n(x)[/tex]

[tex]P_n(-x) = (-1)^n P_n(x)[/tex]

[tex]P_n'(-x) = -(-1)^n P_n'(x)[/tex]

And started the job. Now...

[tex][(1-x^2)P_n'(x)]' = -n(n+1)P_n(x)[/tex]

multiplying both sides with [tex]P_m(x)[/tex] and integrating from -1 to 1

[tex]\int_{-1}^{1} [(1-x^2)P_n'(x)]' P_m(x) dx = \int_{-1}^{1} -n(n+1)P_n(x) P_m(x) dx[/tex]

applying a partial to the lefthand side

[tex](1-x^2)P_n'(x) P_m(x) |_{-1}^{1} - \int_{-1}^{1} (1-x^2)P_n'(x) P_m'(x) dx = \int_{-1}^{1} -n(n+1)P_n(x) P_m(x) dx[/tex]

switching n and m's, ve get another equation like it

[tex](1-x^2)P_m'(x) P_n(x) |_{-1}^{1} - \int_{-1}^{1} (1-x^2)P_m'(x) P_n'(x) dx = \int_{-1}^{1} -m(m+1)P_m(x) P_n(x) dx[/tex]

substituing these

[tex][-n(n+1) + m(m+1)]\int_{-1}^{1} P_n(x) P_m(x) dx = [(1-x^2)(P_n'(x) P_m(x) - P_m'(x) P_n(x))] |_{-1}^{1}[/tex]

now, from the inventory

[tex]P_n(-x)P_m'(-x) = -P_n(x) P_m'(x)[/tex]

the righthand side becomes

[tex]\lim_{x \to 1} 2[(1-x^2)(P_n'(x) P_m(x) - P_m'(x) P_n(x))[/tex]

and in it's final shape

[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = \lim_{x \to 1} 2(1-x^2) \frac {(P_n'(x) P_m(x) - P_m'(x) P_n(x))}{-n(n+1) + m(m+1)}[/tex]

For [tex]n \neq m[/tex], there's no problem, it's straightly 0. But for [tex]n = m[/tex]...

And everything starts going wrong. By stating

[tex](1-x^2)P_n'(x) = \int -n(n+1)P_n(x) dx[/tex]

i'm making a little change

[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = \lim_{x \to 1} 2 \frac {P_m(x) (\int -n(n+1)P_n(x) dx) - P_n(x) (\int -m(m+1)P_m(x) dx))}{-n(n+1) + m(m+1)}[/tex]

Now I plug [tex]P_n(x) = P_m(x)[/tex], [tex]x=1[/tex] and [tex]P_n(1) = 1[/tex] and state

[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = \lim_{x \to 1} 2 \frac {P_m(x) (\int -n(n+1)P_n(x) dx) - P_n(x) (\int -m(m+1)P_m(x) dx))}{-n(n+1) + m(m+1)}[/tex]

[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = 2 (\int P_n(x) dx)_{x=1}[/tex]

which is the nonsense of the day.

Obviously something is wrong. But where's the error?

Thanks in advance.

P.S.: I made an awful mistake https://www.physicsforums.com/showthread.php?p=1092694" in math forum, where I was welcomed with the words "utter and complete nonsense" with no further actual explanation. Well, back to home!

The book says:

[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = \delta_{mn} \frac{2}{2n+1}[/tex]

So I sat and tried derieving it. First, I gather an inventory that might be useful:

[tex](1-x^2)P_n''(x) - 2xP_n'(x) + n(n+1) = 0[/tex]

[tex][(1-x^2)P_n'(x)]' = -n(n+1)P_n(x)[/tex]

[tex]P_n(-x) = (-1)^n P_n(x)[/tex]

[tex]P_n'(-x) = -(-1)^n P_n'(x)[/tex]

And started the job. Now...

[tex][(1-x^2)P_n'(x)]' = -n(n+1)P_n(x)[/tex]

multiplying both sides with [tex]P_m(x)[/tex] and integrating from -1 to 1

[tex]\int_{-1}^{1} [(1-x^2)P_n'(x)]' P_m(x) dx = \int_{-1}^{1} -n(n+1)P_n(x) P_m(x) dx[/tex]

applying a partial to the lefthand side

[tex](1-x^2)P_n'(x) P_m(x) |_{-1}^{1} - \int_{-1}^{1} (1-x^2)P_n'(x) P_m'(x) dx = \int_{-1}^{1} -n(n+1)P_n(x) P_m(x) dx[/tex]

switching n and m's, ve get another equation like it

[tex](1-x^2)P_m'(x) P_n(x) |_{-1}^{1} - \int_{-1}^{1} (1-x^2)P_m'(x) P_n'(x) dx = \int_{-1}^{1} -m(m+1)P_m(x) P_n(x) dx[/tex]

substituing these

[tex][-n(n+1) + m(m+1)]\int_{-1}^{1} P_n(x) P_m(x) dx = [(1-x^2)(P_n'(x) P_m(x) - P_m'(x) P_n(x))] |_{-1}^{1}[/tex]

now, from the inventory

[tex]P_n(-x)P_m'(-x) = -P_n(x) P_m'(x)[/tex]

the righthand side becomes

[tex]\lim_{x \to 1} 2[(1-x^2)(P_n'(x) P_m(x) - P_m'(x) P_n(x))[/tex]

and in it's final shape

[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = \lim_{x \to 1} 2(1-x^2) \frac {(P_n'(x) P_m(x) - P_m'(x) P_n(x))}{-n(n+1) + m(m+1)}[/tex]

For [tex]n \neq m[/tex], there's no problem, it's straightly 0. But for [tex]n = m[/tex]...

And everything starts going wrong. By stating

[tex](1-x^2)P_n'(x) = \int -n(n+1)P_n(x) dx[/tex]

i'm making a little change

[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = \lim_{x \to 1} 2 \frac {P_m(x) (\int -n(n+1)P_n(x) dx) - P_n(x) (\int -m(m+1)P_m(x) dx))}{-n(n+1) + m(m+1)}[/tex]

Now I plug [tex]P_n(x) = P_m(x)[/tex], [tex]x=1[/tex] and [tex]P_n(1) = 1[/tex] and state

[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = \lim_{x \to 1} 2 \frac {P_m(x) (\int -n(n+1)P_n(x) dx) - P_n(x) (\int -m(m+1)P_m(x) dx))}{-n(n+1) + m(m+1)}[/tex]

[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = 2 (\int P_n(x) dx)_{x=1}[/tex]

which is the nonsense of the day.

Obviously something is wrong. But where's the error?

Thanks in advance.

P.S.: I made an awful mistake https://www.physicsforums.com/showthread.php?p=1092694" in math forum, where I was welcomed with the words "utter and complete nonsense" with no further actual explanation. Well, back to home!

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