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I am trying to evaluate:

$$\int_0^{10} \operatorname{sn}(x\mid i) \, dx$$

where $\operatorname{sn}$ denotes the Jacobi elliptic function sn.

The indefinite integral is:

$$(-1)^{3/4} \tanh ^{-1}\left(\sqrt[4]{-1} \operatorname{cd}(x\mid i)\right).$$

But I cannot substitute the limits to compute the definite integral because there is a discontinuity (of the indefinite integral) along the path around $x=7.07$. The definite integral can be computed if the point of discontinuity can be determined analytically.

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  • $\begingroup$ the integrand does not have a singularity, it is continuous on $0<x<10$ $\endgroup$
    – Carlo Beenakker
    Commented Oct 5 at 15:42
  • $\begingroup$ the integrand is continuous, correct. I meant that the indefinite integral has a discontinuity. $\endgroup$
    – mattTheMathLearner
    Commented Oct 5 at 16:14
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    $\begingroup$ crossposted at math.stackexchange.com/q/4980499/87355 --- it does not seem a trivial question, perhaps keep it open? $\endgroup$
    – Carlo Beenakker
    Commented Oct 5 at 16:39

1 Answer 1

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The indefinite integral $$F(x)=(-1)^{3/4} \operatorname{arctanh}\left[(-1)^{1/4} \operatorname{cd}(x\mid i)\right]$$ of the Jacobi elliptic function $\operatorname{sn}(x\mid i)$ has a discontinuity of $$\delta=2\lim_{z\rightarrow\infty}(-1)^{3/4} \operatorname{arctanh}\left[(-1)^{1/4} z\right]= -\frac{(1+i) \pi }{\sqrt{2}}$$ at $x=7.0748\cdots$ [arising from the pole in $\operatorname{cd}(x\mid i)$], hence the definite integral equals $$\int_0^{10} \operatorname{sn}(x\mid i) \, dx=F(10)-F(0)+\delta.$$ This evaluates to $-2.42813 - 2.31227\, i$, in agreement with a numerical evaluation of the integral.

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  • $\begingroup$ Thanks @CarloBeenakkerr, very helpful indeed $\endgroup$
    – mattTheMathLearner
    Commented Oct 6 at 9:48

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