MHB Evaluating Orthogonal Integral for Legendre Equations

[ognik]

The book states that the integral of the (inner) product of 2 distinct eigenvectors must vanish if they are orthogonal.

Given $ P_1(x)=x, Q_0(x)= \frac{1}{2}\ln\left({\frac{1+x}{1-x}}\right) $ are solutions to Legendres eqtn., evaluate their orthog. integral.

Using $ ln(1-z)=-\sum_{n=0}^{\infty}\frac{1}{n} {z}^{n}, |z|< 1 $

$ \ln\left({1+x}\right)= \sum_{n=0}^{\infty}\frac{1}{n} {(-1)}^{n+1}\left(\frac{1}{x}\right)^{\!{n}} $
$ \ln\left({1-x}\right)= -\sum_{n=0}^{\infty}\frac{1}{n} \left(\frac{1}{x}\right)^{\!{n}} $

I got $ \ln\left({\frac{1+x}{1-x}}\right)=-\sum_{n=0}^{\infty}\frac{1}{n}\left(\frac{1}{x}\right)^{\!{n}} $

x is its own series, so $ \int_{-1}^{1} \frac{x}{2} \ln\left({\frac{1+x}{1-x}}\right) \,dx = - \int_{-1}^{1} \frac{x}{2}\sum_{n=0}^{\infty}\frac{1}{n}\left(\frac{1}{x}\right)^{\!{n}}\,dx =
- \frac{1}{2} \int_{-1}^{1} \sum_{n=0}^{\infty}\frac{1}{n}{x}^{-n+1}\,dx $

$ = - \frac{1}{2} \sum_{n=0}^{\infty}\frac{1}{2-{n}^{2}}{x}^{-n+2}|^{1}_{-1}
= \frac{1}{2} \sum_{n=0}^{\infty}\frac{1}{2-{n}^{2}} ({x}^{3}-x)$

I would appreciate confirmation all the above is correct?

So the orthogonal integral is not = 0, except for x=0, therefore the solutions are not orthogonal, the integral must =0 for all values in the interval.

Finding it = 0 at x=0 troubles me, is this significant in some way?

 

[Ackbach]

Can't you just use properties of the logarithm? I would think something more like
$$\int_{-1}^1 x \cdot \frac{1}{2}\ln\left({\frac{1+x}{1-x}}\right) dx=\frac12 \int_{-1}^1 x\left[\ln(1+x)-\ln(1-x)\right]=\frac12 \left[\int_{-1}^1 x\ln(1+x) \, dx-\int_{-1}^1 x\ln(1-x) \, dx \right]$$
might be a bit easier to compute, and less error-prone.

 

[ognik]

Indeed it will, and to over-complicate my life further I had used Laurent series when a simple taylor series would have done - so please don't anyone waste time checking my series version.

 

[ognik]

Having said that, it was very fiddly and I ended with $ \frac{1}{4}[(x^2-1)ln\frac{1+x}{1-x}+2x] $ I was quite pleased with myself until I checked it with Mathematica - $ -\frac{1}{2} Ln (-5 + 4 x + x^2 + 4 Log[-1 + x]) $
Please check my 1st part:

$ \int x ln(1+x) = ln(1+x)\frac{x^2}{2} - \int \frac{x^2}{2} \frac{1}{1+x} = \frac{1}{2}[ ln(1+x)x^2-\int (x-1) - \int \frac{1}{1+x}]$

$ = \frac{1}{2}[{x}^{2} ln(1+x) - \frac{(x-1)^2}{2} - ln(1+x)] = \frac{1}{2}[({x}^{2}-1) ln(1+x) - \frac{(x-1)^2}{2}]$

 

[Ackbach]

Hmm.
\begin{align*}
\frac12 \left[\int_{-1}^1 x\ln(1+x) \, dx-\int_{-1}^1 x\ln(1-x) \, dx \right]
&=\frac12 \left[\int_{0}^2 (u-1)\ln(u) \, du-\int_{0}^2 (1-u)\ln(u) \, du \right] \\
&=\int_{0}^{2}(u-1) \ln(u) \, du \\
&=\int_0^2 u \ln(u) \, du -\int_0^2 \ln(u) \, du \\
&=\left(\ln(u)\cdot\frac{u^2}{2}\right)\Bigg|_{0}^{2}-\int_0^2 \frac{u^2}{2}\cdot\frac1y \, dy-\left[ \left(\ln(u) \cdot u\right)|_{0}^{2}-\int_0^2 y\cdot\frac1y \, dy \right]\\
&=2\ln(2)-\lim_{u\to 0^{+}}\frac{u^2\ln(u)}{2}-1-\left[ 2\ln(2) -\lim_{u\to 0^{+}}(u \ln(u))-2 \right]\\
&=1.
\end{align*}
This doesn't look right, eh? It is confirmed by Mathematica, though. Indeed, the command

Integrate[LegendreP[1,x] * LegendreQ[0,x],{x,-1,1}]

yields $1$, as above. Apparently, these two functions are not orthogonal. I'll have to think about why that is.

 

[ognik]

. Apparently, these two functions are not orthogonal. I'll have to think about why that is.

It was expected that they would not be orthogonal, the next question asks me to explain why 'the proof of orthogonality does not apply'

- - - Updated - - -

While I really like the elegance of your solution, and will work to remember what makes it elegant, did you happen to spot any errors in my effort?

 

[ognik]

Hi ackbach (or anyone)...

So the 2 solutions are not orthogonal and fail the orthogonality integral test, fine. But the Legendre eqtn is self-adjoint, so the eigenvectors must be orthogonal. I think the answer is in the interval - Legendre is bounded within [1,-1] therefore the solutions must also be bounded in that interval, but clearly $Q_0$ is not bounded at either end point...
I think this right, but there is probably more that could be said - for example are they valid solutions in this case?

Please check my 1st part:

$ \int x ln(1+x) = ln(1+x)\frac{x^2}{2} - \int \frac{x^2}{2} \frac{1}{1+x} = \frac{1}{2}[ ln(1+x)x^2-\int (x-1) - \int \frac{1}{1+x}]$

$ = \frac{1}{2}[{x}^{2} ln(1+x) - \frac{(x-1)^2}{2} - ln(1+x)] = \frac{1}{2}[({x}^{2}-1) ln(1+x) - \frac{(x-1)^2}{2}]$

I'd really appreciate putting this to rest, is my working for $ \int x ln(1+x)$ correct, if not please point out the error (so I don't do it again later). Perhaps my solution and Ackbach's are equivalent? I didn't like in the mathematica solution, having ln(...log[..]) - does this appear in maths? Why do they use ln and log, I thought ln was natural log, and lof was base 10?

 

[Ackbach]

I think your antiderivative is correct. You can always check it by differentiating, and see if you get the original integrand. Somehow the definite integral became an indefinite integral - you need your limits there.

As for ln versus log, you have to understand that Mathematica was created by Stephen Wolfram, a physicist. So, ln, as a mathematical standard, is replaced in Mathematica by Log. It's a Mathematica thing, and you just have to get used to it. If it really bothers you, then in each notebook, you can execute ln[x_]=Log[x], and then use ln[x] to your heart's content.