14
$\begingroup$

The complete elliptic integral of the first kind is defined as $$K(k)=\int_0^{\pi/2} \frac{dx}{\sqrt{1-k^2\sin^2{x}}}$$ and the complete elliptic integral of the second kind is defined as $$E(k)=\int_0^{\pi/2}\sqrt{1-k^2\sin^2{x}}~dx$$ for $0\leq k<1$.

I'm supposed to prove the following relation $$K'(k)=\frac{E(k)}{k(1-k^2)}-\frac{K(k)}{k}.$$

What I tried so far

Without much thought about the exchange of integration and differentiation I tried to compute \begin{align}K'(k)&=\int_0^{\pi/2} \frac{k\sin^2{x}}{(1-k^2\sin^2{x})^{3/2}}dx=-\frac{1}{k}\int_0^{\pi/2}\left(\frac{1-k^2\sin^2{x}}{(1-k^2\sin^2{x})^{3/2}}-\frac{1}{(1-k^2\sin^2{x})^{3/2}}\right)dx\\ &=-\frac{K(k)}{k}+\frac{1}{k}\int_0^{\pi/2}\frac{1}{(1-k^2\sin^2{x})^{3/2}} dx.\end{align} Comparing this with the result I'm supposed to obtain it would remain to show $$\int_0^{\pi/2}\frac{1}{(1-k^2\sin^2{x})^{3/2}} dx=\int_0^{\pi/2}\frac{\sqrt{1-k^2\sin^2{x}}}{1-k^2}dx.$$ Some numerical computations suggest that this identity is correct but I have know idea how to show it. Any hints or solutions would be appreciated!

$\endgroup$
10
+50
$\begingroup$

Let us make the change of variables $t=\sin^2x$, $\displaystyle dx=\frac{dt}{2\sqrt{t(1-t)}}$ so that \begin{align} &K(k)=\frac12\int_0^{1}\frac{dt}{\sqrt{t(1-t)(1-k^2 t)}},\tag{1}\\ &E(k)=\frac12\int_0^{1}\sqrt{\frac{1-k^2t}{t(1-t)}}\,dt=\frac12\int_0^{1}\frac{(1-k^2t)dt}{\sqrt{t(1-t)(1-k^2 t)}}\,dt.\tag{2} \end{align} Then, using (1) and (2), we deduce that \begin{align} &k(1-k^2)K'(k)-E(k)+(1-k^2)K(k)=\\ &=\frac12\int_0^1\frac{k^2(1-k^2)t\,dt}{(1-k^2t)\sqrt{t(1-t)(1-k^2 t)}}-\frac12\int_0^1\frac{k^2(1-t)dt}{\sqrt{t(1-t)(1-k^2 t)}}=\\ &=-k^2\int_0^{1}\frac{d}{dt}\left(\sqrt{\frac{t(1-t)}{1-k^2t}}\,\right)dt=\\ &=0. \end{align} At the last step we integrated the derivative of a function vanishing at the endpoints.


P.S. If instead you would like to show your last identity, the idea remains the same: use that $$\frac{1}{(1-k^2\sin^2x)^{3/2}}-\frac{(1-k^2\sin^2x)^{1/2}}{1-k^2}=-\frac{k^2}{1-k^2} \frac{d}{dx}\left(\frac{\sin x\cos x}{\sqrt{1-k^2\sin^2x}}\right).$$

$\endgroup$

protected by J. M. is a poor mathematician May 23 '17 at 6:44

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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