# Piecewise continuous contours with discontinuity only at end points

Let $w(t)=u(t)+iv(t)$ where $a \leq t \leq b$ be a complex valued function on real variable $t$. For integrating $w(t)$ from $a$ to $b$ we require that $u(t)$ and $v(t)$ must be piecewise continuous.According to Brown and Churchill piecewise continuous if it is continuous everywhere in the stated interval except possibly for a finite number of points.But at these points one sided limits must exist.If discontinuity is at end points only one of the one sided limit (right hand limit at $a$ ,left hand limit at $b$) is required.

Now let $u(t), a \leq t \leq b$ be a function which is continuous everywhere except at $a$. As per definition right hand limit at $a$ exist.Then there are two possibilities One is $u(a)$ is not defined.But this can be ruled out by a previous discussion in stack exchange.

Second one is $u(a)$ is defined but not equal to the limit.

My problem is that how to find $\int_{a}^{b} u(t)dt$ in such case . For example if my $u(t)$ is defined in such a way that

$u(t)=2 \quad t=0$

$u(t)=t$, $0 <t\leq 1$

Then what will be $\int_{0}^{1} u(t)dt$ ?

Caveat: for contour integration you are normally interested a complex-valued function $\gamma$ on a closed interval $[a,b]$ in the reals, whose derivative $\gamma'$ is piecewise continuous because what you want to calculate is $\int_a^b f(\gamma(t))\gamma'(t)\,dt$.
Corrected: This reminds me a lot to the definition of the Dirac delta, for some strange reason. Try defining: $$u_a(t) = 2e^{-at} + t$$ Note that: $$\lim\limits_{a\to\infty}u_a(t) = t$$ But at the same time: $$\forall a\in\mathbb{R}:\quad u_a(0) = 2$$ The best part is that $u_a(t)$ is integrable in $0\leq t\leq1$. Thus: $$\int\limits_{0}^{1} u_a(t) dt = \int\limits_{0}^{1} 2e^{-at} + t dt = \dfrac{4+a-4e^{-a}}{2a}$$ Taking limit as $a \to \infty$: $$\lim\limits_{a \to \infty} \dfrac{4+a-4e^{-a}}{2a} = \dfrac{1}{2}$$ So: $$\int\limits_{0}^{1} t dt = \dfrac{1}{2} = 0.5 \quad\quad\therefore\quad \int\limits_{0}^{1} t dt = \lim\limits_{a \to \infty} \int\limits_{0}^{1} u_a(t) dt$$