Piecewise continuous contours with discontinuity only at end points

Question

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$ ?

Answer 1

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: $$ \int\limits_{0}^{1} t dt = \dfrac{1}{2} = 0.5 \quad\quad\therefore\quad \int\limits_{0}^{1} t dt = \lim\limits_{a \to \infty} \int\limits_{0}^{1} u_a(t) dt $$

Answer 2

I don't know the Brown and Churchill text. However, in the more usual integration of complex-valued functions over real intervals discussed in Baby Rudin, changing the value of the function being integrated at just one point does not change the integral.

For Riemann integrals (ch. 6) this can be seen by refining partitions by adding an interval of small size surrounding the point in question and noting that the contribution of the interval to upper and lower Riemann sums for those refined partitions goes to zero as the length of the interval goes to zero. This reasoning closely follows that found in the proof of proposition 6.10 in Rudin, which tells us that a bounded function continuous except at a finite number of points has a Riemann integral.

For Lebesgue integrals (ch. 11) it's even easier to see: one point has measure zero, and the Lebesgue integral does not change if the integrand is changed on a set of Lebesgue measure zero.

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 $\int_a^b f(\gamma(t))\gamma'(t)\,dt$.