Convergence of a complete elliptic integral

Question

How to prove that the following integral assuming $k^2 < 1$ $$\int_{0}^{1} \frac{d x}{\sqrt{1-x^2}\sqrt{1-k^2 x^2}}$$ is convergent?

Assuming[k^2 < 1, 
 Integrate[1/Sqrt[(1 - x^2)*(1 - k^2*x^2)], {x, 0, 1}]]

EllipticK[k^2]

FunctionRange[{EllipticK[k^2], 0 <= k^2 < 1}, k, y, Reals]

y >= Pi/2

It can also be expressed as how to prove convergence of the following elliptic integral?

EllipticK[k^2], (k^2<1)

Answer 1

In fact the original question asks for demonstrating that the following integral is finite, what Matematica can do simply

FullSimplify[ Integrate[1/(Sqrt[1 - x^2] Sqrt[1 - k^2 x^2]), {x, 0, 1}]  
              < Infinity,
             -1 < k < 1 ]

True

A mathematical proof is comparably easy. We can see that for $\; -1< k<1$ and $\;0\leq x <1$ we have $\frac{1}{ \sqrt{1-k^2 x^2}} \leq \frac{1}{\sqrt{1-k^2}}\;$ and so $$\frac{1}{\sqrt{1-x^2} \sqrt{1-k^2 x^2}} \leq \frac{1}{\sqrt{1-k^2}\sqrt{1-x^2}}$$ and since the both sides of this inequality are nonnegative, integration is a generalization of summation we get $$\int_{0}^{1}\frac{dx}{\sqrt{1-x^2} \sqrt{1-k^2 x^2}} \leq \int_{0}^{1}\frac{dx}{\sqrt{1-k^2}\sqrt{1-x^2}} =\frac{\pi}{2\sqrt{1-k^2}}$$ recalling that $\int_{0}^{1}\frac{dx}{\sqrt{1-x^2}}=\frac{\pi}{2}$

Q.E.D.

A counterpart of a crucial step in the proof can be demonstrated simply as well

Simplify[ 1/(Sqrt[1 - x^2] Sqrt[1 - k^2 x^2]) 
          <= 1/(Sqrt[1 - x^2] Sqrt[1 - k^2]), 
                      -1 < k < 1 && 0 <= x < 1]

True

We've got a better upper bound for the complete elliptic integral of the first kind namely $\frac{\pi}{2 \sqrt{1-k^2}}$. Analogously we can find lower bound of the form $\frac{\pi \arcsin(k)}{2 k}$ and in a slightly different way another lower bound $\frac{\log(\frac{1-k}{1+k})}{2 k}$. The first one works for $x$ close to $0$ and the other one for $x$ close to $1$.

We show all the graphs of appropriate functions on the following plot

Plot[{ EllipticK[k^2], Pi/(2 Sqrt[1 - k^2]), Pi/2 (ArcSin[k]/k), 
      Log[(1 + k)/(1 - k)]/(2 k)}, {k, 0, 1}, 
  PlotStyle -> {Thick, Dashed, Dashed, Dashed}, AxesOrigin -> {0, 0}, 
  PlotLegends -> "Expressions", PlotRange -> {0, 5}]

Answer 2

Integrate[1/Sqrt[(1 - x^2)*(1 - k^2*x^2)], {x, 0, 1}]

ConditionalExpression[EllipticK[k^2], Re[k^2] <= 1 || k^2 ∉ Reals]

FunctionDomain[EllipticK[k^2], k]

-1 < k < 1

The definition domain just indicated that the EllipticK[k^2] converge only in -1 < k < 1 and divergence outside such interval.