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I'm currently trying to solve the sum $$ f(x)=\sum\limits_{n=0}^\infty\frac{x^{n+1}}{n+1}P_n(x), $$ where $P_n(x)$ is the Legendre function of order n.

I also named the sum $f(x)$ since I'm the solution will be a function of $x$. The first thing that can be noticed is that the function $f(x)$ is an uneven function.

My first step was to take the derivative of $f(x)$, this then yields that: $$ f'(x)=\sum\limits_{n=0}^\infty x^{n}P_n(x)+\sum\limits_{n=1}^\infty\frac{x^{n+1}}{n+1}P'_n(x). $$ Using the generating function $$ \frac{1}{\sqrt{1-2hz+h^2}}=\sum\limits_{n=0}^\infty h^nP_n(z) $$ I'm able to rewrite the first term as: $$ f'(x)=\frac{1}{\sqrt{1-x^2}}+\sum\limits_{n=1}^\infty\frac{x^{n+1}}{n+1}P'_n(x). $$ Now for the second term I'm still stuck, I've been trying a few recursive relations, in the hope to maybe get a first order differential equation for $f(x)$. So far I haven't found the right recursive relation yet.

Now I'm wondering if there is maybe a better approach that I should have taken, or that if there is an easy trick to solve the second term?

After playing with the function in Mathematica I figured out that the most probable solution is $$ f(x)=\mathrm{arctanh}(x), $$ this gives me a direction, but still I have no clue for the second term. All help is welcome!

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  • $\begingroup$ When I plot the sum of the first 20 terms it looks more like $\arcsin{x}$, and of course, you obtained $f(x)=\arcsin{x}+ \int\sum\limits_{n=1}^\infty\frac{x^{n+1}}{n+1}P'_n(x)$. $\endgroup$
    – krvolok
    Commented Aug 10, 2015 at 16:05
  • $\begingroup$ @krvolok, my bad I meant the arctanh. I also considered the arcsin, but it did not fit. Not enough bend and wrong values. $\endgroup$
    – Nick
    Commented Aug 11, 2015 at 5:41

2 Answers 2

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Let: $$f(x) = \sum _{n=0}^{\infty} \frac{x^{n+1} \, P_{n}(x)}{n+1}$$ As you said: $$f^{'}(x) = \sum_{n=0}^{\infty} x^{n} \, P_{n}(x) + \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} \, P_{n}^{'}(x)$$ And using the generating function you showed, we get: $$f^{'}(x) = \frac{1}{\sqrt{1-x^{2}}} + \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1 } \, P_{n}^{'}(x)$$ Now, note that: $$ { P }_{ n }(x)'\quad =\quad \frac { \left( n+1 \right) \left( x{ P }_{ n }(x)\quad -\quad { P }_{ n+1 }(x) \right) }{ 1-{ x }^{ 2 } } $$

So we get: \begin{align} f^{'}(x) &=\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } +\sum _{ n=0 }^{ \infty }{ \frac { { x }^{ n+1 } }{ n+1 } } \frac { \left( n+1 \right) \left( x{ P }_{ n }(x) - { P }_{ n+1 }(x) \right) }{ 1-{ x }^{ 2 } } \\ &= \frac {1}{\sqrt{1-x^{2}}} + \frac{1}{1-x^{2}} \sum_{n=0}^{\infty}{x^{n+1} \left(x {P}_{n}(x) - {P}_{n+1}(x) \right)} \\ &= \frac{1}{\sqrt{1-x^{2}}} + \frac{1}{1-x^{2}} \sum_{n=0}^{\infty} x^{n+2} \left( P_{n}(x) \right) - \frac{1}{1-x^{2}} \sum_{n=0}^{\infty} x^{n+1} \, P_{n+1}(x) \\ &= \frac{1}{\sqrt{1-x^{2}}} +\frac{x^{2}}{1-x^{2}} \sum_{n=0}^{\infty} x^{n} \left(P_{n}(x) \right) - \frac{1}{1-x^{2}}\left(\sum_{n=0}^{\infty} x^{n} \, P_{n}(x) -1 \right) \\ &= \frac{1}{\sqrt{1-x^{2}}} + \frac{x^{2}}{1-x^{2}} \frac{1}{\sqrt{1-x^{2}}} - \frac{1}{1-x^{2}} \left(\frac{1}{\sqrt{1-x^{2}}} -1\right) \\ &= \frac{1}{\sqrt{1-x^{2}}} + \frac{x^{2}-1}{ (1-x^{2}) \, \sqrt{1-x^{2}}} + \frac{1}{1-x^{2}} \\ &= \frac{1}{\sqrt{1-x^{2}}} + \frac{-1}{\sqrt{1-x^{2}}} + \frac{1}{1-x^{2}} = \frac {1}{1-x^{2}} \end{align} Integration yields: \begin{align} f(x) &= \int \frac{dx}{1-x^{2}} = \frac{1}{2} \, \int \left(\frac{1}{1-x} + \frac{1}{1+x} \right) dx \\ &= \frac{1}{2} \, \left[\ln(1+x)-\ln(1-x)\right] + C \\ &= \frac{\ln\left(\frac{1+x}{1-x}\right)}{2} + C \end{align} and for $x=0$ we have $f(0) = 0$ so we get $C=0$.

Hence our answer is: $$f(x) = \sum_{n=0}^{\infty}{\frac{{x}^{n+1}}{n+1} \, P_{n}(x)} = \frac{1}{2} \, \ln\left(\frac{1+x}{1-x}\right)$$

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  • $\begingroup$ great! But I think your reccurence relation might be wrong. Maybe an error with the summation index ? $\endgroup$
    – Nick
    Commented Aug 11, 2015 at 6:02
  • $\begingroup$ I don't think it is wrong, numerical approach agrees(try for multiple values of $x$ and compare you'll see ti's correct) $\endgroup$ Commented Aug 11, 2015 at 11:01
  • $\begingroup$ @Dominique Note that $\frac{\ln{(\frac{1+x}{1-x})}}{2} = arctanh(x)$ $\endgroup$ Commented Aug 11, 2015 at 11:45
  • $\begingroup$ yes with the difference that you have $n+1$ as index on your last term while wikipedia has $n-1$. That was also the reason I was thinking of summation indices because the $n-1$ is only possible when you sum from $n=1$, not $n=0$. And indeed I saw that de logarithm and arctangent coincide :). $\endgroup$
    – Nick
    Commented Aug 11, 2015 at 13:08
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    $\begingroup$ I didn't check the relation for the derivative, but I'll just accept it for what it is :p. The solution fits with numerics anyway. $\endgroup$
    – Nick
    Commented Aug 12, 2015 at 14:31
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While the other response is a good exercise in exploiting recurrence relations, a much more direct route is provided by the technique of summing under the integral which converts the series into an integral of the generating function:

$$\begin{align} f{\left(x\right)} &=\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}P_{n}{\left(x\right)}\\ &=\sum_{n=0}^{\infty}P_{n}{\left(x\right)}\int_{0}^{x}\mathrm{d}t\,t^{n}\\ &=\int_{0}^{x}\mathrm{d}t\,\sum_{n=0}^{\infty}t^{n}P_{n}{\left(x\right)}\\ &=\int_{0}^{x}\mathrm{d}t\,\frac{1}{\sqrt{1-2xt+t^2}}\\ &=\int_{-x}^{\frac{\sqrt{1-x^2}-1}{x}}\mathrm{d}u\,\frac{2}{1-u^2};~~~\small{\left[\sqrt{1-2xt+t^2}=tu+1\right]}\\ &=2\int_{\frac{1-\sqrt{1-x^2}}{x}}^{x}\frac{\mathrm{d}w}{1-w^2};~~~\small{\left[-u=w\right]}\\ &=2\operatorname{arctanh}{\left(x\right)}-2\operatorname{arctanh}{\left(\frac{1-\sqrt{1-x^2}}{x}\right)}\\ &=\operatorname{arctanh}{\left(x\right)}.\blacksquare\\ \end{align}$$

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    $\begingroup$ Beautiful method +1 $\endgroup$
    – Autolatry
    Commented Aug 11, 2015 at 7:55

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