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Category:Software distributionQ:

Inconsistent proof of an equation involving multiple integrals

Let $F(z)=\int_0^z f(t)\,\mathrm{d}t$, where $f$ is a complex polynomial such that $f(0)=\frac{1}{4}$ and $f(\mathrm{i})=\mathrm{i}$
I have to show that $F(\mathrm{i})=-\mathrm{i}$. I did it like this:
$F(z)=\int_0^z f(t)\,\mathrm{d}t$
$F(\mathrm{i})=\int_0^\mathrm{i}f(t)\,\mathrm{d}t+\int_\mathrm{i}^\mathrm{i}f(t)\,\mathrm{d}t$
$F(\mathrm{i})=\int_0^\mathrm{i}\overline{f(\mathrm{i})}+f(\mathrm{i})\,\mathrm{d}t$
$F(\mathrm{i})=2\mathrm{i}+\overline{\mathrm{i}}=\mathrm{i}-\mathrm{i}=0$
I can see that $F(\mathrm{i})=\mathrm{i}$ by considering the function $g(t)=t$, which is such that $g(0)=0$ and $g(\mathrm{i})=\mathrm{i}$
But what I don’t understand is why I can get $F(\mathrm{i})=\mathrm{i}$ for $F(z)=\int_0^z f(t)\,\mathrm{d}t$ and then $F(\mathrm{i})=-\mathrm{i}$ for $F(z)=\int_0^\mathrm{i}f(t)\,\mathrm{d}t+\int_\mathrm{i}^\mathrm{i}f(t)\,\mathrm{d}t$

A:

The two integrals are different.
The first one is a linear integral starting
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