Integration over a Real Interval

Let $f(t) = u(t) + iv(t)$ for $a \le t \le b$ , where $u$ and $v$ are real-valued continuous functions of the real variable $t$ .

\int_a^b f(t)\,\text{d}t = \int_a^b u(t)\,\text{d}t + i\int_a^b v(t)\,\text{d}t

If $U'(t) = u(t)$ and $V'(t) = v(t)$ for each $a \le t \le b$ ,

\int_a^b f(t)\,\text{d}t = (U(b) - U(a)) + i(V(b) - V(a))

Properties

Let $f(t) = u(t) + iv(t)$ and $g(t) = p(t) + iq(t)$ be continuous on $[a,b]$ .

\int_a^b (f(t) + g(t))\,\text{d}t = \int_a^b f(t)\,\text{d}t + \int_a^b g(t)\,\text{d}t

\int_a^b f(t)\,\text{d}t = \int_a^c f(t)\,\text{d}t + \int_c^b f(t)\,\text{d}t, \qquad a \le c \le b

Let $\alpha = r + is$ be a complex constant.

\int_a^b \alpha f(t)\,\text{d}t = \alpha \int_a^b f(t)\,\text{d}t

\int_a^b f(t)\,\text{d}t = -\int_b^a f(t)\,\text{d}t

Let $f$ be continuously differentiable on $[a,b]$ .

\int_a^b f'(t)\,\text{d}t = f(b) - f(a)

\int_a^b f(t)g(t)\,\text{d}t = \int_a^b (u p - v q)\,\text{d}t + i\int_a^b (u q + v p)\,\text{d}t

Evaluate $\displaystyle\int_0^1 (1 + it)^2\,\text{d}t$ .

Solution Reveal Hide

Expand the integrand.

(1 + it)^2 = 1 + 2it + i^2 t^2 = (1 - t^2) + i(2t)

So $u(t) = 1 - t^2$ and $v(t) = 2t$ .

Integrate the real part.

\int_0^1 (1 - t^2)\,\text{d}t = \left[t - \frac{t^3}{3}\right]_0^1 = 1 - \frac{1}{3} = \frac{2}{3}

Integrate the imaginary part.

\int_0^1 2t\,\text{d}t = \left[t^2\right]_0^1 = 1

Combine using the definition.

\int_0^1 (1 + it)^2\,\text{d}t = \frac{2}{3} + i

Evaluate $\displaystyle\int_0^1 (t - i)^3\,\text{d}t$ .

Solution Reveal Hide

Expand the integrand.

(t - i)^3 = t^3 - 3it^2 + 3i^2 t - i^3 = (t^3 - 3t) + i(1 - 3t^2)

Integrate the real part.

\int_0^1 (t^3 - 3t)\,\text{d}t = \left[\frac{t^4}{4} - \frac{3t^2}{2}\right]_0^1 = \frac{1}{4} - \frac{3}{2} = -\frac{5}{4}

Integrate the imaginary part.

\int_0^1 (1 - 3t^2)\,\text{d}t = \left[t - t^3\right]_0^1 = 0

Combine using the definition.

\int_0^1 (t - i)^3\,\text{d}t = -\frac{5}{4}

Evaluate $\displaystyle\int_0^{\pi/2} e^{t + it}\,\text{d}t$ .

Solution Reveal Hide

Write the integrand as $e^{(1+i)t}$ . An antiderivative is

\frac{e^{(1+i)t}}{1+i} = \frac{(1-i)\,e^{(1+i)t}}{2}

Evaluate at the limits, using $e^{(1+i)\pi/2} = ie^{\pi/2}$ .

Solution Reveal Hide

\frac{(1-i)\,ie^{\pi/2}}{2} - \frac{1-i}{2} = \frac{(1+i)\,e^{\pi/2}}{2} - \frac{1-i}{2}

Separate the real and imaginary parts.

\int_0^{\pi/2} e^{t + it}\,\text{d}t = \frac{1}{2}\left(e^{\pi/2} - 1\right) + \frac{i}{2}\left(e^{\pi/2} + 1\right)