2
$\begingroup$

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

$\endgroup$
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.