Calculating $$I=\int_{-1}^1{\dfrac{1}{\sqrt{1-x}}P_n(x)} \, dx$$ Where $P_n$ is a Legendre Polynomial.

My progress: For any integral of the form: $$\int_{-1}^1{f(x)P_n(x)} \, dx$$ Usinng Rodrigues formula, and integrating by parts $n$-times like in this example, we can assure that: $$\int_{-1}^1{f(x)P_n(x)} \, dx=\dfrac{(-1)^n}{2^n n!} \int_{-1}^1 \frac{d^n}{dx^n} \big(f(x)\big)\cdot (x^2-1)^n $$

So, in this case: $f(x)=(1-x)^{-1/2}$

$$ I=\int_{-1}^1{\dfrac{1}{\sqrt{1-x}}P_n(x)} \, dx = \frac{(-1)^n}{2^n n!} \int_{-1}^1\frac{d^n}{dx^n} \big( (1-x)^{-1/2} \big) \cdot (x^2-1)^n \, dx$$

Where: $$\frac{d^n}{dx^n} \big( (1-x)^{-1/2} \big)=(-1)^n (1-x)^{-1/2-n}(1/2-n)_n$$ Using Pochhammer symbols.


$$I=\frac{(1/2-n)_n}{2^n n!} \int_{-1}^1 (1-x)^{-1/2-n}\cdot (x^2-1)^n \, dx$$

And here is where i have problems, sometimes, Betta function araises, but i can´t see how.

  • 2
    $\begingroup$ Hint: $\frac{1}{\sqrt{1-x}}=\sqrt{2}\sum_{k\geq0}P_k(x)$. Now use the orthogonality of the L egendre polynomials! $\endgroup$
    – tired
    Jan 13, 2016 at 20:14
  • 1
    $\begingroup$ tired's good hint makes the problem very easy to solve. Depending on how you define $P_k$ you should probably prove it. If you use the generating function definition then it's trivial otherwise $\frac{1}{\sqrt{1-2xt+t^2}} = \sum_{k\geq 0}P_k(x)t^k$ can be shown from the recursion formula: multiply the rec. formula by $t^{n+1}$ and sum over $n=1,2,3,\ldots$. Then recognize one side as the derivative of the other to get a simple ODE for the function $\sum_{k\geq 0}P_k(x)t^k$. $\endgroup$
    – Winther
    Jan 13, 2016 at 20:44

1 Answer 1


As noted in the comments by tired the problem is easy once you establish $$\frac{1}{\sqrt{1-x}} = \sqrt{2}\sum_{n=0}^\infty P_n(x)\tag{1}$$ With this in hand the ortogonality relation $\int_{-1}^1 P_m(x)P_n(x){\rm dx} = \frac{2\delta_{nm}}{2m+1}$ does the rest of the job $$\color{red}{\int_{-1}^1\frac{P_m(x)}{\sqrt{1-x}}{\rm d}x = \sqrt{2}\sum_{n=0}^\infty \int_{-1}^1P_n(x)P_m(x){\rm d}x = \frac{2\sqrt{2}}{2m+1}}$$ Below I will give a derivation of $(1)$ in case you haven't see it before.

$P_k(x)$ turns out to be the coefficient of $t^k$ in the generating function $$G(x,t) =\sum_{n=0}^\infty P_n(x) t^n = \frac{1}{\sqrt{1-2xt+t^2}}$$ and by taking $t=1$ we recover $(1)$.

If you don't know the generating function this can be derived either from Legendre's differential equation, from Rodrigues' formula using the Leibniz rule or from the recursion formula $$(n+1)P_{n+1}(x) = (2n+1)xP_n(x) - nP_{n-1}(x)$$ I will here give a derivation based on this latter definition.

Multiply the recursion formula by $t^{n+1}$ and sum over $n=1,2,3\ldots$ to get $$\left[\sum_{n=0}^\infty nP_n(x)t^{n-1}\right](1-2xt+t^2) = (x-t)\left[\sum_{n=0}^\infty P_n(x)t^n\right]$$ We recognize the left hand side as the derivative $\frac{d}{dt}\left[\sum_{n=0}^\infty P_n(x)t^n\right] \equiv \frac{dG(x,t)}{dt}$ giving us the ODE $$\frac{dG(x,t)}{dt} = \frac{(x-t)}{1-2xt+t^2}G(x,t) \implies G(x,t) = e^{\int_0^t\frac{x-t'}{1-2xt'+t'^2}{\rm d}t'} = \frac{1}{\sqrt{1-2xt+t^2}}$$ where I used the substitution $u =1-2xt'+t'^2 \implies {\rm d}u = -2(x-t'){\rm d}t'$ to evaluate the integral.

  • 1
    $\begingroup$ I'm glad that u made such a nice answer out of my hint. A big (+1)! $\endgroup$
    – tired
    Jan 14, 2016 at 14:19

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