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I have this continuous function $f:\mathbb{C}\rightarrow\mathbb{C}$ defined on an open set $\Omega$.
I also have a family of identical smooth curves up to translation $z_{t}:[a,b]\rightarrow\mathbb{C}$ in $\Omega$ and $t \in [0, T]$, where the parameter $t$ specifies some linear translation, i.e. $z_{t+\delta} = z_{t} + \delta K$ where $K$ is some constant. (You can imagine that the curve $z_{t}$ is "shifting" with the passage of time, $t$.)

So my question is, is the integral $\int_{z_{t}}f(x) dx$ continuous as a function of $t$? In other words, as my curve $z_{t}$ moves a little bit in $\Omega$, does the value of the integral also move a little bit?

More specifically, in my case I know that the value of the integral is $0$ for all $t \in (\alpha, \beta]$. Must the value of the integral also be zero for $t=\alpha$?

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Write down the definitions: $\int_{z_t} f(x)dx = \int_a^b f(z_t(s))\frac{d}{ ds} z_t(s) ds$ With your 'translation' property of the $z_t$ the derivative wrt $s$ does not depend on $t$, so only the $f(z_t(s))$ plays a role. Now write down the difference for $t = \alpha$ and $t$ close to $\alpha$ and estimate the difference: $\int_a^b \left(f(z_\alpha(s)) - f(z_t(s))\right)\frac{d}{ ds} z_t(s) ds $ Since the image of $[0,T]\times [a,b]$ under $z$ is compact und $f$ continuous, $f$ is uniformly continuous on that image, hence the difference can be made uniformly arbitrarily small as $t\rightarrow \alpha$, given your definition of $z_t$.

(This reasoning, of course, relies on the assumption that the images of all $z_t$ are contained in $\Omega$ and do not, e.g. tend to the boundary of $\Omega$ as $t\rightarrow \alpha$. But since you've chosen closed intervals as domain of definition, I assume this to be true.)

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    Yes, of course, I forgot to norm the derivatives. Thanks a bunch! If I fix the errors, your comment will no longer make sense so I guess I'll leave it.2012-06-04