# Convergence of a complete elliptic integral

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)

• Integrate[1/Sqrt[(1 - x^2)*(1 - k^2*x^2)], {x, 0, 1}] Jan 31, 2022 at 2:20
• FunctionDomain[EllipticK[k^2], k] Jan 31, 2022 at 2:21
• It is convergent because there is only one singularity on the path, at the endpoint x=1, and it is easuly seen to be integrable by series expansion: In[188]:= Series[1/Sqrt[(1 - x^2) (1 - k^2 x^2)], {x, 1, 1}, Assumptions -> -1 < k < 1] Out[188]= SeriesData[x, 1, { 2^Rational[-1, 2] (1 - k^2)^Rational[-1, 2] (1 - x)^Rational[-1, 2], Rational[-1, 4] 2^Rational[-1, 2] ( 1 - k^2)^Rational[-1, 2] (-1 + k^2)^(-1) (-1 + 5 k^2) ( 1 - x)^Rational[-1, 2]}, 0, 2, 1] Jan 31, 2022 at 14:08

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}]


• Thank you very much! Feb 1, 2022 at 2:46
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.

• That is not so simple in view of FunctionDomain[EllipticK[k^2], k, Complexes] resulting in -1 + k^2 != 0. It's unclear to me whether the integral Integrate[1/Sqrt[(1 - x^2)*(1 - k^2*x^2)], {x, 0, 1}] converges for such values of k. Jan 31, 2022 at 8:33
• ClearAll[k]; Integrate[1/Sqrt[(1 - x^2)*(1 - k^2*x^2)], {x, 0, 1}, Assumptions -> k \[Element] Reals && (k^2 > 1 || k^2 < -1)] outputs (-I EllipticK[1 - 1/k^2] + EllipticK[1/k^2])/Abs[k]. Jan 31, 2022 at 9:59
• Thank you!@cvgmt@user64494 Feb 1, 2022 at 2:48
• EllipticK[k^2] does converge for $k>1$ contrary to your statment: "only in -1 < k < 1 and ..." Feb 24, 2023 at 11:46