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Given the the Legendre polynomials generating function:

$$G(x,t)=\frac{1}{\sqrt{1-2xt+t^2}}=\sum_{n=0}^{\infty}P_n(x)t^n$$

prove the relation:

$$\int_{-1}^{1} (P_n(x))^2 dx = \frac{2}{2n+1}$$

My attempt:

We can say that:

$$G(x,t)G(x,u)=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}P_n(x)P_m(x)t^nu^m=\frac{1}{\sqrt{1-2xt+t^2}}\frac{1}{\sqrt{1-2xu+u^2}}$$

And argue that $\int_{-1}^{1}G(x,t)G(x,u)dx$ is a function of $tu$ and not upon $t$ and $u$ individually because all $\int_{-1}^{1}P_n(x)P_m(x)dx=0$. That means that only the terms with $n=m$ in the double summation survive.

Now, I guess you should perform:

$$\int_{-1}^{1}G(x,t)G(x,u)dx=\int_{-1}^{1}\frac{1}{\sqrt{1-2xt+t^2}}\frac{1}{\sqrt{1-2xu+u^2}}dx$$

But the integral ends up being unbelievably complicated and long.

Is it right to assume $t=u$? The book where this comes from says that the answer is:

$$\frac{1}{z}ln\frac{1+z}{1-z}$$ $$z = (tu)^{1/2}$$

Any help/pointing would be appreciated.

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  • $\begingroup$ Is it possible to use $$\int_{-1}^{1}\sum_{0}^{\infty} (tu)^nP_n^2(x)dx$$? If it is, how can it be used? $\endgroup$ – Federico Lopez Oct 20 '19 at 9:59
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We know,from generating function of Legendre polynomials,

$(1-2xz+z^2)^\frac{-1}{2}=\sum z^nP_n(x)$

Squaring both sides and writing the sum by separating the cross terms and "self-quadrature" terms,we get

$(1-2xz+z^2)^{-1}=\sum z^{2n}P_n^2(x)+2\sum z^{m+n}P_m(x)P_n(x)$

Integrating both sides between -1 and +1,we have,

$\int_{-1}^1\sum z^{2n}P_n^2(x)dx+\int_{-1}^1 2\sum z^{m+n}P_m(x)P_n(x)dx=\int_{-1}^{1}(1-2xz+z^2)^{-1}dx$

Now,

$\int_{-1}^{1}P_n(x)P_m(x)dx=0$ due to orthogonality of Legendre polynomials and the 2nd integral on LHS of the above equation is wrt x,the terms containing m and n can be taken out,and therefore the 2nd integral vanishes.

$\Rightarrow \int_{-1}^1\sum z^{2n}P_n^2(x)dx+0=\int_{-1}^{1}(1-2xz+z^2)^{-1}dx$

The integral on RHS,i.e., $ {\displaystyle\int}\dfrac{1}{-2zx+z^2+1}\,\mathrm{d}x $ can be solved as--

Substitute $u=-2zx+z^2+1$

$\dfrac{\mathrm{d}u}{\mathrm{d}x} = -2z$ $\Rightarrow \mathrm{d}x=-\dfrac{1}{2z}\,\mathrm{d}u$

The integral reduces to $=-\class{steps-node}{\cssId{steps-node-1}{\dfrac{1}{2z}}}{\displaystyle\int}\dfrac{1}{u}\,\mathrm{d}u$ which is a standard integral.

Therefore the integral becomes $=-\dfrac{\ln\left(u\right)}{2z}$ $=-\dfrac{\ln\left(-2zx+z^2+1\right)}{2z}$

$\Rightarrow \sum z^{2n}\int_{-1}^1P_n^2(x)dx=\frac{-1}{2z}[\ln(1-2xz+z^2)]_{-1}^1$

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

Using Maclaurin Series of $\ln(1+z)$ and $\ln(1-z)$ and expanding,

RHS= $\frac{1}{z}[(z-\frac{z^2}{2}+\frac{z^3}{3}-\frac{z^4}{4}+...)-(-z-\frac{z^2}{2}-\frac{z^3}{3}-\frac{z^4}{4}-...)]$

Observe that the terms containing even powers of x cancel and the odd terms gets doubled(due to presence of them in both the series indicated above).Taking the 1/z term inside the square brackets,

$\Rightarrow \int_{-1}^1\sum z^{2n}P_n^2(x)dx=2[1+\frac{z^2}{3}+\frac{z^4}{5}+..\frac{z^{2n}}{2n+1}.]$

Finally,equating the coefficients of $z^{2n}$ on both sides we have,

$\int_{-1}^{1} (P_n(x))^2 dx = \frac{2}{2n+1}$,which is the required relation.

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  • $\begingroup$ All the summations have index running from zero to infinity. $\endgroup$ – Manas Dogra Oct 29 '19 at 12:24

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