# summation of Legendre polynomials over a power law

I am trying to find a closed-form for the summation: $$\sum_{n=0}^{\infty}\frac{P_{n}(x)}{(n+k)^{\alpha}}$$ where $$P_{n}(x)$$ denote the Legendre polynomials, k is a constant, and $$\alpha$$ is positive integer. I have seen some special cases here for $$\alpha=1$$, But I am really interested when $$\alpha$$ is 2 or 3 or even higher. I do not know how to implement generating function for the Legendre polynomials here. Thank you.

Note that this is a solution as an integral, not a closed form solution

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

Multiply by $$t^{k-1}$$ and get

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

We replace $$t\mapsto u$$ and integrate $$u$$ from $$0$$ to $$t$$ and get

$$\int_0^t\frac{u^{k-1}}{\sqrt{1-2xu+u^2}}\,du=\sum_{n=0}^\infty\frac{P_n(x)t^{n+k}}{n+k}$$

Now divide by $$t$$, do another substitution $$t\mapsto v$$ and integrate this from $$0$$ to $$t$$ to get

$$\int_0^t\frac{1}{v}\int_0^v\frac{u^{k-1}}{\sqrt{1-2xt+t^2}}\,dudv=\sum_{n=0}^\infty\frac{P_n(x)t^{n+k}}{(n+k)^2}$$

This double integral is performed over a triangular region in $$uv$$ space. Switching the order of integration yields

$$\int_0^t\int_u^t\frac{1}{v}\frac{u^{k-1}}{\sqrt{1-2xt+t^2}}\,dvdu=\sum_{n=0}^\infty\frac{P_n(x)t^{n+k}}{(n+k)^2}$$

We can evaluate the integral in $$v$$ to get

$$\int_0^t\frac{u^{k-1}\ln(t/u)}{\sqrt{1-2xt+t^2}}\,du=\sum_{n=0}^\infty\frac{P_n(x)t^{n+k}}{(n+k)^2}$$

We continue this process, but the result is weird. We again divide by $$t$$, replace $$t\mapsto v$$, and integrate from $$0$$ to $$t$$ to get

$$\int_0^t\frac{1}{v}\int_0^v\frac{u^{k-1}\ln(v/u)}{\sqrt{1-2xu+u^2}}\,dudv=\sum_{n=0}^\infty\frac{P_n(x)t^{n+k}}{(n+k)^3}$$

We swap the order and factor some terms to make the equation a bit simpler to get

$$\int_0^t\left(\int_u^t\frac{\ln(v/u)}{v}\,dv\right)\frac{u^{k-1}}{\sqrt{1-2xu+u^2}}\,du=\sum_{n=0}^\infty\frac{P_n(x)t^{n+k}}{(n+k)^3}$$

We do a change of variable; let $$w=v/u$$ and $$dw=(1/u)dv$$ to get

$$\int_0^t\left(\int_1^{t/u}\frac{\ln(w)}{w}\,dw\right)\frac{u^{k-1}}{\sqrt{1-2xu+u^2}}\,du=\sum_{n=0}^\infty\frac{P_n(x)t^{n+k}}{(n+k)^3}$$

Integrating with respect to $$w$$ gives

$$\frac{1}{2}\int_0^t\frac{u^{k-1}\ln^2(t/u)}{\sqrt{1-2xu+u^2}}\,du=\sum_{n=0}^\infty\frac{P_n(x)t^{n+k}}{(n+k)^3}$$

This process can be repeated indefinitely (I'll skip the rigorous details, since they are long and tedious) and we are left with the generalized equation of

$$\frac{1}{(\alpha-1)!}\int_0^t\frac{u^{k-1}\ln^{\alpha-1}(t/u)}{\sqrt{1-2xu+u^2}}=\sum_{n=0}^\infty\frac{P_n(x)t^{n+k}}{(n+k)^\alpha},\;\;\;\;\alpha=1,2,3,...$$

Let $$t=1$$ and we have

$$\boxed{ \frac{(-1)^{\alpha-1}}{(\alpha-1)!}\int_0^1\frac{u^{k-1}\ln^{\alpha-1}(u)}{\sqrt{1-2xu+u^2}}=\sum_{n=0}^\infty\frac{P_n(x)}{(n+k)^\alpha},\;\;\;\;\alpha=1,2,3,... }$$

We can see how a solution for $$\alpha=1$$ could be obtained, since the log term disappears in this form. A more straightforward approach to this answer comes from a well known result of

$$\int_0^1 t^\mu\ln^k(t)\,dt=(-1)^k\frac{k!}{(\mu+1)^{k+1}}$$

Thus, we could have simply multiplied the generating function by $$t^{k-1}\ln^{\alpha-1}(t)$$, integrated from $$0$$ to $$1$$, then rearranged a bit.