Let $c > 0$ and let the function $f : (0, \infty) \to \mathbb{C}$ be defined as $$ f(y) = \int_{c - i\infty}^{c+i\infty} \frac{y^s}{s(s+1)} \, ds. $$ I want to show that $f$ is continuous.
My attempt: For $T > 0$ we define $$ I(y, T) = \int_{c-iT}^{c+iT} \frac{y^s}{s(s+1)} \, ds. $$ Then by the triangle inequality we have \begin{align} \left| f(y) - I(y, T) \right| &\leq \left| \int_{c-i\infty}^{c-iT} \frac{y^s}{s(s+1)} \, ds \right| + \left| \int_{c+iT}^{c+i\infty} \frac{y^s}{s(s+1)} \, ds \right| \\ &\leq 2y^c \int_{T}^{\infty} \frac{1}{t^2} \, dt. \end{align} Let $M > 0$. Then note that $I(y, t)$ converges uniformly to $f(y)$ in the open interval $0 < y < M$. Thus $f$ is continuous in the interval $(0, M)$ since $I(y, T)$ is continuous as a function of $y$. As $M$ was taken to be arbitrary, it follows that $f$ is continuous.
My Question: I want to ask whether the above proof is an overkill, i.e., whether there is a way to recognize if such an integral defines a continuous function by just looking at it.
