# Equality of integrals

this is q.2 of ahlfors p. 241: Show that $$\int_{-1}^{1}\frac{dt}{\sqrt {(1-t^2)(1-k^2t^2)}}=\int_{1}^{\frac{1}{k}}\frac{dt}{\sqrt {(t^2-1)(1-k^2t^2)}}$$ if and only if $k=(\sqrt{2}-1)^2$ . Thank you.

• Lars (I assumed you meant Lars) Ahlfors wrote several books. Perhaps you should specify which one. – Willie Wong Apr 26 '13 at 14:31

This is basically an exercise in elliptic function theory. The integral on the left is clearly equal to $\displaystyle I_{1} = 2\int_{0}^{1}\frac{dt}{\sqrt{(1 - t^{2})(1 - k^{2}t^{2})}} = 2\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1 - k^{2}\sin^{2}\theta}} = 2K(k)$

The integral on right is bit tricky. But we have to see that by definition of elliptic function $\text{sn}(u, k)$ we have $\displaystyle u = \int_{0}^{\text{sn}(u, k)}\frac{dt}{\sqrt{(1 - t^{2})(1 - k^{2}t^{2})}}$ and we need to use the knowledge that $\text{sn}(K + iK', k) = 1/k$ so that $\displaystyle K + iK' = \int_{0}^{1/k}\frac{dt}{\sqrt{(1 - t^{2})(1 - k^{2}t^{2})}}$

Then we can see that

$\displaystyle K + iK' = \int_{0}^{1}\frac{dt}{\sqrt{(1 - t^{2})(1 - k^{2}t^{2})}} + \int_{1}^{1/k}\frac{dt}{\sqrt{(1 - t^{2})(1 - k^{2}t^{2})}}$ $\displaystyle \Rightarrow iK' = \int_{1}^{1/k}\frac{dt}{\sqrt{(1 - t^{2})(1 - k^{2}t^{2})}}$

$\displaystyle \Rightarrow K' = \int_{1}^{1/k}\frac{dt}{\sqrt{(t^{2} - 1)(1 - k^{2}t^{2})}}$

So your original equation between integrals becomes $K'(k) = 2K(k)$ or $K'(k)/K(k) = \sqrt{4}$. Now clearly this means that $k$ is a singular modulus $k_{n}$ with $n = 4$. The value of $k = k_{4}$ is a standard one given by $k = (\sqrt{2} - 1)^{2}$.

A complete understanding would require one to read material on elliptic functions, class invariants and singular moduli. Some of this material is presented in my blog : http://paramanands.blogspot.com/2011/01/elliptic-functions-introduction.html http://paramanands.blogspot.com/2012/03/ramanujans-class-invariants.html

To obtain the value of $k$ we can use the Landen's transformation (http://paramanands.blogspot.com/2009/08/pi-and-the-agm-evaluating-elliptic-integrals.html) directly to get $$K(k) = \frac{1}{1 + k}K\left(\frac{2\sqrt{k}}{1 + k}\right)$$ and if we put $l = 2\sqrt{k}/(1 + k)$ we can see that $k = (1 - l')/(1 + l')$ so that we express the same relation as $$K(l) = \frac{2}{1 + l'}K(k)$$ and we can also see that $k' = 2\sqrt{l'}/(1 + l')$ and hence if we replace $k$ by $l'$ in first relation above we get $$K(l') = \frac{1}{1 + l'}K\left(\frac{2\sqrt{l'}}{1 + l'}\right) = \frac{1}{1 + l'}K(k')$$

On division we now obtain $$\frac{K(k')}{K(k)} = 2\frac{K(l')}{K(l)}$$ where $k = (1 - l')/(1 + l')$. If we put $l = 1/\sqrt{2}$ so that $l' = \sqrt{1 - l^{2}} = 1/\sqrt{2} = l$ then we see that we have $$\frac{K(k')}{K(k)} = 2$$ where $$k = \dfrac{1 - \dfrac{1}{\sqrt{2}}}{1 + \dfrac{1}{\sqrt{2}}} = (\sqrt{2} - 1)^{2}$$