$\lim_{y\rightarrow 0^{+}}(f(x+iy)-f(x-iy))$ I was trying to solve this exercize.
Let $\varphi:[0,1]\rightarrow\mathbb{C}$ be a continuous function and 
$$f(z)=\int_{0}^1 \dfrac{\varphi(t)}{t-z} dt, z\in\mathbb{C}\setminus[0,1].$$
I was able to prove that $f$ is holomorphic and $$f^{(k)}(z)=k!\int_{0}^1 \dfrac{\varphi(t)}{(t-z)^{k+1}} dt$$
I want to find $$\lim_{y\rightarrow 0^{+}}(f(x+iy)-f(x-iy)),$$ for $x\in[0,1]$. I guess that the limit should be an expression of $\varphi(x)$, but apart from that I have no idea even how to start. Any ideas?
 A: To supplement C.Dubussy's answer, note that we have
$$f(x+iy)-f(x-iy)=\int_0^1 \frac{2iy\phi(t)}{(t-x)^2+y^2}\,dt$$
For $x\in (0,1)$, fix a number $\delta>0$ so that $\delta <x <1-\delta$.  Then,  write
$$\begin{align}
f(x+iy)-f(x-iy)&=\int_0^{x-\delta} \frac{2iy\phi(t)}{(t-x)^2+y^2}\,dt\\\\
&+\int_{x-\delta}^{x+\delta} \frac{2iy\phi(t)}{(t-x)^2+y^2}\,dt\\\\
&+\int_{x+\delta}^{1} \frac{2iy\phi(t)}{(t-x)^2+y^2}\,dt \tag 1
\end{align}$$
As $y\to 0^+$, the first and third integrals on the right-hand side of $(1)$ vanish.  We analyze next the second integral on the right-hand side of $(1)$.  We have
$$\begin{align}
\int_{x-\delta}^{x+\delta} \frac{2iy\phi(t)}{(t-x)^2+y^2}\,dt&=\int_{-\delta}^{\delta} \frac{2iy\phi(t+x)}{t^2+y^2}\,dt\\\\
&=\int_{-\delta}^{\delta} \frac{2iy(\phi(t+x)-\phi(x))}{t^2+y^2}\,dt+2i\phi(x)\int_{-\delta}^\delta \frac{y}{t^2+y^2}\,dt\\\\
&=\int_{-\delta}^{\delta} \frac{2iy(\phi(t+x)-\phi(x))}{t^2+y^2}\,dt+4i\phi(x)\arctan(\delta/y) \tag 2
\end{align}$$
By continuity of $\phi(t)$, for any $\epsilon >0$, we can find a $\delta>0$ so that the magnitude of integral in $(2)$ is smaller than $\epsilon$.  Upon passing to the limit as $y \to 0^+$, we find for $x\in (0,1)$
$$\bbox[5px,border:2px solid #C0A000]{\lim_{y\to 0^+}(f(x+iy)-f(x-iy))=2\pi i \phi(x)}$$

NOTE:
If $x=0$ or $x=1$, then the previous development can be slightly modified to show
$$\bbox[5px,border:2px solid #C0A000]{\lim_{y\to 0^+}(f(x+iy)-f(x-iy))=\pi i \phi(x)}$$
