# Prove that the Dirac delta function can be presented in terms of the Legendre Polynomials

I tried everything I could, but I can't find a derivation for this equation. Can I get some hints on how to approach to prove the following result? $$\delta(1-x)=\sum_{n=0}^{\infty}{\frac{(2n+1)P_n(x)}{2}}.$$

• Please give more details about the problem, where you got it from, what you thought about it, where you got stuck, and you can write many other things as well. There is no word limit for your questions. It helps the community understand what exactly you need, and helps them be able to answer you more accordingly. May 12 at 17:20
• How is your formula consistent with formula (41) at mathworld.wolfram.com/LegendrePolynomial.html which indicates $\left(1-2 x t+t^2\right)^{-3/2} \left[\left(2 x t-2 t^2\right)+\left(1-2 t x+t^2\right)\right]=\sum\limits_{n=0}^\infty (2 n+1)\, P_n(x)\, t^n$? May 12 at 23:55
• @StevenClark why do you think the two equations are inconsistent or even related for that matter. The OP equation is a special case of formula 22 at dlmf.nist.gov/1.17 . May 14 at 5:33
• @BillWatts I don't know. I'm not sure about the range of convergence, but setting $t=1$ leads to $0=\sum\limits_{n=0}^\infty (2 n+1)\, P_n(x)$ which I suppose is consistent with $\delta(1-x)=\sum_{n=0}^\infty{\frac{(2n+1)P_n(x)}{2}}$ everywhere except at $x=1$. When I plot $\sum_{n=0}^N{\frac{(2n+1)P_n(x)}{2}}$ the sum seems to diverge to infinity as the evaluation limit $N$ increases for negative $x$ as well as positive $x$, so I'm wondering over what range of values of $x$ the sum could represent $\delta(1-z)$. May 14 at 14:41

Just create a Fourier series using the orthogonality of the Legendre Polynomials. Set

$$\delta (1-x)=\sum _{n=0}^{\infty } A_n\ P_n(x)$$

Multiply both sides by $$P_{n1}(x)$$ and integrate.

$$\int_{-1}^1 \delta (1-x) P_{n1}(x) \, dx=\sum _{n=0}^{\infty } A_n \int_{-1}^1 P_n(x) P_{{n1}}(x) \, dx$$

$$1$$=$$2$$ $$\frac{A_n \delta _{\text{n1},n}}{2 n+1}$$

And solve for $$A_n$$.

• Your integration limits (-1,1) would work for an integral over $\delta(x)$, but not necessarily for an integral over $\delta(1-x)$. May 12 at 22:38
• @StevenClark They work as long as the user specifies that $1$ is within the limits of (-1,1). If the user specifies $1$ to be outside those limits, then of course the integral over the $\delta$ is $0$. May 12 at 22:44