I have encountered the following relationship$^{[1][2]}$, stated without proof both times

$$\int_0^\gamma dt \tan(t/2)\cdot [P_n(\cos(t))+P_{n-1}(\cos(t))]=\frac{1}{n}[P_{n-1}(\cos(\gamma))-P_{n}(\cos(\gamma))] $$

Where $P_n(x)$ is the $n^{th}$ Legendre polynomial, and $\gamma<\pi$.

I've tried integration by parts, substitution of $u=\cos(t)$, and looked in Gradshteyn. Mathematica doesn't integrate it (at least, not in the current form). I've verified that the relationship is correct for modest values of $n$. I suspect that there is some property of the Legendre polynomials that facilitates this, but I don't see what it is.

[1] Mixed boundary value problems, Dean G. Duffy, eq. 3.1.80, p95

[2] This paper, eq. A8


1 Answer 1


Since $\tan(x/2)=\frac{\sin(x)}{1+\cos(x)}$, by letting $\cos(\gamma)=T\in[-1,1]$ your identity can be written in the simplified form

$$ \int_{T}^{1}\frac{dx}{1+x}(P_n(x)+P_{n-1}(x))\,dx = \frac{1}{n}(P_{n-1}(T)-P_n(T)). $$

The identity clearly holds at $T=1$. By differentiating both sides wrt $T$ we find Bonnet's identity and we are done.


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