I am struggling to understand some of the proof that $\displaystyle\int_{x=-1}^{1}[P_L(x)]^2\,\mathrm{d}x=\frac{2}{2L+1}\tag{1}$

In my book I have a list of $6$ recursion relations for Legendre Polynomials which I will show for reference below, as one of them is needed for the proof:

$(a)\quad LP_L(x)=(2L-1)xP_{L-1}(x)-(L-1)P_{L-2}(x)$

$(b)\quad xP_L\acute(x)-P_{L-1}\acute(x)=LP_L(x)$

$(c)\quad P_L\acute(x)-xP_{L-1}\acute(x)=LP_{L-1}(x)$

$(d)\quad (1-x^2)P_L\acute(x)=LP_{L-1}(x)-LxP_L(x)$

$(e)\quad (2L+1)P_L(x)=P_{L+1}\acute(x)-P_{L-1}\acute(x)$

$(f)\quad (1-x^2)P_{L-1}\acute(x)=LxP_{L-1}(x)-LP_L(x)$

The book proof goes as follows:

To prove $(1)$ we use recursion relation $(b)$, namely, $$LP_L(x)=xP_L\acute(x)-P_{L-1}\acute(x)\tag{2}$$ Multiply $(2)$ by $P_L(x)$ and integrate to get $$L\int_{x=-1}^{1}[P_L(x)]^2\,\mathrm{d}x=\int_{x=-1}^{1}xP_L(x)P_L\acute(x)\,\mathrm{d}x-\color{blue}{\int_{x=-1}^{1}P_L(x)P_{L-1}\acute(x)\,\mathrm{d}x}\tag{3}$$ The last ($\color{blue}{\mathrm{blue}}$) integral is zero by Problem 7.4. To evaluate the middle integral in $(3)$ we integrate by parts $$\int_{x=-1}^{1}xP_L(x)P_L\acute(x)\,\mathrm{d}x= \frac{x}{\color{red}{2}}[P_L(x)]^2\Big |_{x=-1}^{1}-\frac{1}{\color{red}{2}}\int_{x=-1}^{1}[P_L(x)]^2\,\mathrm{d}x\tag{4}$$

I do not understand why there is a $\color{red}{2}$ in the denominators on the RHS of $(4)$.

Is anyone able to explain how the author arrived at $(4)$? Or some help or hints is greatly appreciated. Thank you.

  • 1
    $\begingroup$ The $2$'s come from the antiderivative of $P_L(x)P_L'(x)$ (i.e., $\frac{[P_L(x)]^2}{2}$) in the integration by parts formula: Let $u= x$ and $dv = P_L(x)P_L'(x)\, dx$. $\endgroup$ – kobe Apr 18 '16 at 2:32
  • $\begingroup$ @kobe Thanks. Is there a reason why I cannot let $u=xP_L(x)$ and $dv=P_L'(x)\, dx$? As that substitution will lead to $(4)$ but without the 2's. $\endgroup$ – BLAZE Apr 18 '16 at 2:40
  • 1
    $\begingroup$ Even with your substitution, you would eventually get the $2$'s. I could post up an answer if you like. $\endgroup$ – kobe Apr 18 '16 at 2:45

If you let $u = x$ and $dv = P_L(x)P_L'(x)\, dx$ in the integration by parts formula, you'll obtain $(4)$ since $v = \frac{[P_L(x)]^2}{2}$. Alternatively, you may set $u = xP_L(x)$ and $dv = P_L'(x)\, dx$, so then

\begin{align}\int_{-1}^1 xP_L(x)P_L'(x)\, dx &= x[P_L(x)]^2\bigg|_{-1}^1 - \int_{-1}^1 P_L(x) (P_L(x) + xP_L'(x))\, dx\\ &=x[P_L(x)]^2\bigg|_{-1}^1 - \int_{-1}^1 [P_L(x)]^2\, dx - \int_{-1}^1 xP_L(x)P_L'(x)\, dx\end{align}


$$2\int_{-1}^1 xP_L(x)P_L'(x)\, dx = x[P_L(x)]^2\bigg|_{-1}^1 - \int_{-1}^1 [P_L(x)]^2\, dx$$

Dividing through by $2$ yields $(4)$.

  • $\begingroup$ Thank you for taking the time to explain this to me, it's appreciated. $\endgroup$ – BLAZE Apr 18 '16 at 16:29

It is also possible to do this the hard way, using Rodrigues' formula and 'simply' solving the integral. See Legendre Polynomials: proofs


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