How does one prove the following integral identity, where $P_n(x)$ is the $n$th Legendre polynomial? $$ \int_0^1 P_n(2t^2-1) dt = \frac{(-1)^n}{2n+1} $$

Notes & Background

  • A variant of this appears in, for instance, Erdelyi et al "Higher transcendental functions" 10.10(49), but with nothing in the way of explanation.

  • This comes up in harmonic analysis on $U(3)$, when comparing Gelfand-Tseltin bases associated to different choices of nested sequences $U(3) \supset U(2) \supset U(1)$.

  • Eventually, I'll be looking for a $q$-analogue, related to harmonic analysis on $U_q(3)$, so a proof that will transport well would be my true desire.

  • 2
    $\begingroup$ By means of power series, this transforms into something like $\int\limits_0^1\dfrac{1}{\sqrt{1-2\left(2x^2-1\right)t+t^2}}dx=\mathrm{arctan}t$ (note that I have changed your $t$ into an $x$). Maybe it's easier this way? $\endgroup$ – darij grinberg Apr 21 '11 at 9:30
  • 2
    $\begingroup$ $$\int_0^1 \frac 1 {\sqrt{1-2(2x^2-1)t^2+t^4}}dx = \frac {\arctan t} t$$ might work even better. $\endgroup$ – user11235 Apr 21 '11 at 10:30

Nice idea. As far as I'm concerned, the above comments are "answers", since they check out. I might as well record the details: $$ \int_0^1 \frac{dx}{\sqrt{1-2(2x^2-1)t + t^2}} = \frac{1}{2\sqrt{t}} \int_0^1 \frac{dx}{\sqrt{\frac{(1+t)^2}{4t} -x^2}} = \frac{1}{2\sqrt{t}}\arcsin\left(\frac{\scriptstyle 2\sqrt{t}}{\scriptstyle1+t}\right). $$ The half-angle formula for $\sin$ reduces this to $$ \frac{1}{\sqrt{t}} \arcsin \sqrt{\frac{t}{1+t}} = \frac{1}{\sqrt{t}} \arctan \sqrt{t}, $$ which has the desired power series expansion.


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