I want to show that $$\lim_{\delta \to 0}\int_{\ln(1+\delta)}^{\delta} \frac{e^{-x}}{x}\,dx \to 0 \tag{1}$$
Intutitively I could take the limits before integration, then I would get
$$\int_{0}^{0} \frac{e^{-x}}{x}\,dx=0 $$ My question is whether this procedure is valid, or a more rigorous proof should be provided in he lines of
Edit: Being more careful in the calculations I could show the following estimate:
$$ \begin{align*} \lim_{\delta \to 0}\int_{\ln(1+\delta)}^{\delta} \frac{e^{-x}}{x}\,dx & \leq \left| \lim_{\delta \to 0}\int_{\ln(1+\delta)}^{\delta} \frac{e^{-x}}{x}\,dx \right|\\ & \leq \lim_{\delta \to 0}\int_{\ln(1+\delta)}^{\delta}\left| \frac{e^{-x}}{x}\right|\,dx \\ & \leq \lim_{\delta \to 0}\int_{\ln(1+\delta)}^{\delta} \frac{dx}{x} \\ &=\lim_{\delta \to 0} \, \ln(x)\Big|_{\ln(1+\delta)}^{\delta}\\ &=-\lim_{\delta \to 0} \, \ln\left(\frac{\ln(1+\delta)}{\delta}\right)\\ &=- \ln\left(\lim_{\delta \to 0}\frac{\ln(1+\delta)}{\delta}\right) & \text{by continuity of log}\\ &=- \ln\left(\lim_{\delta \to 0}\frac{1}{\delta}\sum_{n=1}^\infty \frac{(-1)^{n+1} \delta^n}{n}\right)\\ &=- \ln\left(\lim_{\delta \to 0}\sum_{n=1}^\infty \frac{(-1)^{n-1} \delta^{n-1}}{n}\right)\\ &=-\ln(1)\\ &=0 \end{align*} $$
Motivation:
The motivation behind this limit comes from a proof of an integral representation of the Digamma function.
If we define the Gamma function by the integral $(2)$ below and the digamma function by $\psi(z)=\frac{z}{dz}\ln\left(\Gamma(z) \right)$
$$\Gamma(z)= \int_0^\infty e^{-t}t^{z-1}\,dt \qquad \operatorname{Re}(z)>0\tag{2}$$
Differentiating $(2)$ w.r. to $z$ we obtain:
$$ \begin{align*} \Gamma^\prime(z)&= \int_0^\infty e^{-t}t^{z-1} \ln(t)\,dt \qquad \operatorname{Re}(z)>0\\ &= \int_0^\infty e^{-t}t^{z-1} \left(\int_0^\infty \frac{e^{-x}-e^{-xt}}{x}\,dx \right)\,dt\\ &= \int_0^\infty \left(\int_0^\infty (e^{-x}-e^{-xt})e^{-t}t^{z-1}\,dt \right)\,\frac{dx}{x}\\ &= \int_0^\infty \left(e^{-x}\int_0^\infty e^{-t}t^{z-1}\,dt-\int_0^\infty e^{-t(1+x)}t^{z-1}\,dt \right)\,\frac{dx}{x}\\ &= \int_0^\infty \left(e^{-x}\Gamma(z)-\frac{1}{(1+x)^z}\int_0^\infty e^{-t}t^{z-1}\,dt \right)\,\frac{dx}{x}\\ &= \int_0^\infty \left(e^{-x}\Gamma(z)-\frac{1}{(1+x)^z}\Gamma(z)\right)\,\frac{dx}{x}\\ &=\Gamma(z) \int_0^\infty \left(e^{-x}-\frac{1}{(1+x)^z}\right)\,\frac{dx}{x}\\ \end{align*} $$
Therefore we have
$$\psi(z)=\int_0^\infty \left(e^{-x}-\frac{1}{(1+x)^z}\right)\,\frac{dx}{x} \tag{3}$$
Than
$$ \begin{align*} \psi(z)&= \lim_{\delta \to 0}\int_{\delta}^\infty \left(e^{-x}-\frac{1}{(1+x)^z}\right)\,\frac{dx}{x}\\ &= \lim_{\delta \to 0}\left[\int_{\delta}^\infty \frac{e^{-x}}{x}\,dx-\int_{\delta}^\infty \frac{1}{x(1+x)^z}\,dx\right]\\ &= \lim_{\delta \to 0}\left[\int_{\delta}^\infty \frac{e^{-x}}{x}\,dx-\int_{\ln(1+\delta)}^\infty \frac{e^{-xz}}{1-e^{-x}}\,dx\right] & (x+1 \to e^{x})\\ &= \lim_{\delta \to 0}\left[\int_{\ln(1+\delta)}^\infty \frac{e^{-x}}{x}\,dx-\int_{\ln(1+\delta)}^\infty \frac{e^{-xz}}{1-e^{-x}}\,dx+\int_{\delta}^{\ln(1+\delta)} \frac{e^{-x}}{x}\,dx\right] \\ &= \lim_{\delta \to 0}\left[\int_{\ln(1+\delta)}^\infty \frac{e^{-x}}{x}- \frac{e^{-xz}}{1-e^{-x}}\,dx-\int_{\ln(1+\delta)}^{\delta} \frac{e^{-x}}{x}\,dx\right] \\ &=\int_{0}^\infty \frac{e^{-x}}{x}- \frac{e^{-xz}}{1-e^{-x}}\,dx, \qquad \operatorname{Re}(z)>0 \end{align*} $$
Provided $(1)$ holds.
For $\delta\in(0,1)$, we have that \begin{eqnarray*} \ln(1+\delta) & = & \delta-\frac{\delta^{2}}{2}+\frac{\delta^{3}}{3}-\frac{\delta^{4}}{4}+\ldots\\ & \geq & \delta-\frac{\delta^{2}}{2}+\frac{\delta^{3}}{3} \end{eqnarray*} because $-\frac{\delta^{4}}{4}+\frac{\delta^{5}}{5}-\ldots<0$. Since the integrand $x\mapsto\frac{e^{-x}}{x}$ is positive and $\delta-\frac{\delta^{2}}{2}+\frac{\delta^{3}}{3}\leq\ln(1+\delta)\leq\delta$, we have that \begin{eqnarray*} \int_{\ln(1+\delta)}^{\delta}\frac{e^{-x}}{x}dx & \leq & \int_{\delta-\frac{\delta^{2}}{2}+\frac{\delta^{3}}{3}}^{\delta}\frac{e^{-x}}{x}dx\\ & \leq & \int_{\delta-\frac{\delta^{2}}{2}+\frac{\delta^{3}}{3}}^{\delta}\frac{dx}{\delta-\frac{\delta^{2}}{2}+\frac{\delta^{3}}{3}}\\ & = & \frac{1}{\delta-\frac{\delta^{2}}{2}+\frac{\delta^{3}}{3}}\cdot\left(\frac{\delta^{2}}{2}-\frac{\delta^{3}}{3}\right)\\ & \rightarrow & 0 \end{eqnarray*} as $\delta\rightarrow0^{+}$. On the other hand, clearly $\int_{\ln(1+\delta)}^{\delta}\frac{e^{-x}}{x}dx\geq0$. By sandwich rule, we conclude that $\lim_{\delta\rightarrow0+}\int_{\ln(1+\delta)}^{\delta}\frac{e^{-x}}{x}dx=0$.