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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 $$\delta=2\lim_{x\rightarrow\infty}(-1)^{3/4} \operatorname{arctanh}\left[(-1)^{1/4} x\right]= -\frac{(1+i) \pi }{\sqrt{2}}$$ at $x=7.07$ [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.

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.

The indefinite integral $F(x)=(-1)^{3/4} \operatorname{arctanh}\left[(-1)^{1/4} \operatorname{cd}(x\mid i)\right]$ of $f(x)=\operatorname{sn}(x\mid i)$ has a discontinuity of $$\delta=2\lim_{x\rightarrow\infty}(-1)^{3/4} \operatorname{arctanh}\left[(-1)^{1/4} x\right]= -\frac{(1+i) \pi }{\sqrt{2}}$$ at $x=7.07$ [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.

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 $$\delta=2\lim_{x\rightarrow\infty}(-1)^{3/4} \operatorname{arctanh}\left[(-1)^{1/4} x\right]= -\frac{(1+i) \pi }{\sqrt{2}}$$ at $x=7.07$ [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.

the indefinite integral $F(x)=(-1)^{3/4} \tanh ^{-1}\left(\sqrt[4]{-1} \operatorname{cd}(x\mid i)\right)$ of $f(x)=\operatorname{sn}(x\mid i)$ has a discontinuity of $\delta=-\frac{(1+i) \pi }{\sqrt{2}}$ around $x=7.07$, hence the definite integral equals

$$\int_0^{10} \operatorname{sn}(x\mid i) \, dx=f(10)-f(0)+\delta.$$

The indefinite integral $F(x)=(-1)^{3/4} \operatorname{arctanh}\left[(-1)^{1/4} \operatorname{cd}(x\mid i)\right]$ of $f(x)=\operatorname{sn}(x\mid i)$ has a discontinuity of $$\delta=2\lim_{x\rightarrow\infty}(-1)^{3/4} \operatorname{arctanh}\left[(-1)^{1/4} x\right]= -\frac{(1+i) \pi }{\sqrt{2}}$$ at $x=7.07$ [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.