# Under which hypotheses is switching between sum and integral signs legit?

Which hypotheses are needed to change the order of sum and integral signs?

Concrete example: consider the expression $$\int_{\gamma}\frac{f(\zeta)}{\zeta-z_0}\sum_{k=0}^{\infty}\left(\frac{\zeta-z_0}{z-z_0}\right)^k \, d\zeta$$ where $\gamma$ is a circle of radius $r$ of the complex plane. Suppose that the sum $$\sum_{k=0}^{\infty}\left(\frac{\zeta-z_0}{z-z_0}\right)^k$$ is uniformly convergent because $\left|\frac{\zeta-z_0}{z-z_0}\right|<1$. $f$ is holomorphic on a compact set which is a closed annulus of width $R-r$. In this setting, is it sufficient to put the sum sign outside the integral to obtain $$\sum_{k=0}^{\infty} \int_{\gamma}\frac{f(\zeta)}{\zeta-z_0}\left(\frac{\zeta-z_0}{z-z_0}\right)^k \, d\zeta \quad ?$$

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You can consider a summation as an integral w.r.t. the counting measure. Then the Fubini theorem answers your question, see Theorem 2.37 of Folland's book "Real analysis". – Vahid Shirbisheh Feb 9 '13 at 2:42
@VahidShirbisheh: thanks. Nevermind, in the meantime I found what I was looking for: if $\sum_{j=0}^{\infty}f_j(z)$ converges uniformly to $s(z)$ on a domain (i. e. open and connected set) $S$, then for any contour $\gamma$ in $S$ the following holds: $\int_{\gamma}\sum_{j=0}^{\infty}f_j(z)\,\mathrm{d}(z) = \sum_{j=0}^{\infty}\int_{\gamma}f_j(z)\,\mathrm{d}(z)$. – Flast9 Feb 9 '13 at 5:16
Please answer your own question (or ask @VahidShirbisheh to promote the comment to an answer) and accept it to close the issue. Thanks! – vonbrand Feb 9 '13 at 17:09
@vonbrand: let's wait a bit, maybe someone else would add further details. – Flast9 Feb 12 '13 at 6:06