FOUNDATION · CHAPTER 2 OF 8

The integrals, sums and limits that keep coming back

This chapter works through the integrals and sums that signal transforms are built from: exponentials over half-lines, energy and power, swapping a sum and an integral, and the sinc integral. It also explains convergence, so you can tell when a transform exists and when it cannot.

Updated 12 Oct 2026

What you will be able to do

You will evaluate the integral of $e^{- a t}$ over a half-line and say exactly when it converges. You will integrate $t e^{- a t}$ by parts and compute the energy of a signal and the power of a periodic one. You will know when a sum and an integral may be swapped, use the sinc integral, and see why some signals have no transform.

Before you start

Chapter 1, Complex numbers, Euler's formula and the complex exponential, in particular the conjugate $z^{*}$ , $\frac{d}{d t} e^{s t} = s e^{s t}$ for complex $s$ , $∣ e^{j θ} ∣ = 1$ for real $θ$ , and the geometric series with $∣ r ∣ < 1$
Antiderivatives of $e^{- a t}$ and of $1/ t$ (which is $ln t$ ), and substitution in an integral
Integration by parts
Limits at infinity, including $t e^{- a t} \to 0$ for $a > 0$
The identity $cos^{2} θ = \frac{1}{2} (1 + cos 2 θ)$

Exponential integrals over half-lines

Many results in this subject start from one integral: an exponential over a half-line. Continuous time is written $x (t)$ with round brackets and discrete time $x [n]$ with square brackets. The unit step $u (t)$ is 1 for $t > 0$ and 0 for $t < 0$ ; its value at $t = 0$ is left undefined. No result in this chapter depends on it, because changing an integrand at a single point does not change the value of the integral. Everything below rests on one fact: for $a > 0$ , $e^{- a t}$ falls to zero as $t$ grows.

INTEGRAL OF A DECAYING EXPONENTIAL

\int_{0}^{\infty} e^{- a t} d t \int_{- \infty}^{0} e^{a t} d t = [- \frac{e ^{- a t}}{a}]_{0}^{\infty} = 0 - (- \frac{1}{a}) = \frac{1}{a} = [\frac{e ^{a t}}{a}]_{- \infty}^{0} = \frac{1}{a} - 0 = \frac{1}{a}

Both hold for real $a > 0$ , and for complex $a$ with $Re (a) > 0$ . The antiderivative still works for complex $a$ , since $\frac{d}{d t} e^{s t} = s e^{s t}$ (Chapter 1). And $e^{- a t} = e^{- Re (a) t} e^{- j Im (a) t}$ , whose second factor has magnitude 1, so $∣ e^{- a t} ∣ = e^{- Re (a) t} \to 0$ . If $Re (a) \leq 0$ neither converges. $t$ is in seconds, $a$ per second.

WATCH OUT FOR

Writing $1/ a$ without checking the direction. Test: the integrand must shrink as $t$ moves away from the finite limit. $e^{- 2 t}$ shrinks on $[0, \infty)$ , area $\frac{1}{2}$ , but grows on $(- \infty, 0]$ , where the area is infinite.

INTEGRAL OF T TIMES A DECAYING EXPONENTIAL

\int_{0}^{\infty} t e^{- a t} d t = [- \frac{t e ^{- a t}}{a}]_{0}^{\infty} + \frac{1}{a} \int_{0}^{\infty} e^{- a t} d t = 0 + \frac{1}{a} \cdot \frac{1}{a} = \frac{1}{a ^{2}}

Holds for real $a > 0$ , or $Re (a) > 0$ . Integration by parts with $p = t$ , $d q = e^{- a t} d t$ , so $d p = d t$ and $q = - e^{- a t} / a$ . The bracket is 0 at both ends: at $t = 0$ the factor $t$ is 0, and $t e^{- a t} \to 0$ as $t \to \infty$ . With $t$ in seconds the result is in s $^{2}$ .

WORKED EXAMPLE

A signal is $x (t) = 3 t e^{- 2 t} u (t)$ volts, with $t$ in seconds. Find its area $\int_{- \infty}^{\infty} x (t) d t$ .

Solution

The step $u (t)$ is 0 for $t < 0$ , so the integral only runs over $[0, \infty)$ .
Take the constant 3 outside: the area is $3 \int_{0}^{\infty} t e^{- 2 t} d t$ .
Here $a = 2 > 0$ , so the integral converges and equals $1/ a^{2} = 1/4$ .
Multiply: $3 \times \frac{1}{4} = \frac{3}{4}$ .

Answer: The area is $0.75$ V·s.

Energy, power and swapping a sum and an integral

Energy adds up the squared size of the signal over all time. Squaring the magnitude makes every contribution non-negative, so nothing cancels. If the energy is infinite, ask instead for the power $P$ : the average of $∣ x ∣^{2}$ over a long time window. A periodic signal repeats every $T_{0}$ seconds, its period, so one period already gives that average. Sums of many pieces then need a safe way to be integrated piece by piece.

ENERGY OF A SIGNAL

E = \int_{- \infty}^{\infty} ∣ x (t) ∣^{2} d t

For complex signals use $∣ x ∣^{2} = x x^{*}$ . A signal with $0 < E < \infty$ is an energy signal, and its power (below) is 0. For $x$ in volts and $t$ in seconds, $E$ is in V $^{2}$ ·s, which is joules across 1 $Ω$ . In discrete time $E = \sum_{n} ∣ x [n] ∣^{2}$ , summed over all integers $n$ .

POWER OF A SIGNAL AND OF A PERIODIC SIGNAL

P P = T \to \infty lim \frac{1}{2 T} \int_{- T}^{T} ∣ x (t) ∣^{2} d t = \frac{1}{T _{0}} \int_{t_{1}}^{t_{1} + T_{0}} ∣ x (t) ∣^{2} d t (periodic x)

The second form holds only when $x$ repeats with period $T_{0}$ seconds: every period adds the same area, so one period, starting at any time $t_{1}$ , gives the average. The first averages over a window of $2 T$ seconds as $T$ grows. A signal with $0 < P < \infty$ is a power signal; its energy is infinite, so no signal is both. $P$ is in V $^{2}$ , watts across 1 $Ω$ .

WATCH OUT FOR

Thinking a periodic signal has zero energy. If $∣ x ∣^{2}$ has positive area over one period, the energy is infinite. Test: $n$ periods hold $n$ times one period's area, which grows with $n$ , so use power.

WORKED EXAMPLE

Find the energy of $x (t) = 3 e^{- 2 t} u (t)$ volts, with $t$ in seconds.

Solution

Square the magnitude: $∣ x ∣^{2} = 9 e^{- 4 t}$ for $t > 0$ , and 0 for $t < 0$ .
So $E = 9 \int_{0}^{\infty} e^{- 4 t} d t$ .
Apply the half-line integral with $a = 4 > 0$ : it equals $1/4$ .
Multiply: $E = 9/4$ .

Answer: $E = 2.25$ V $^{2}$ ·s, which is finite.

SWAPPING A SUM AND AN INTEGRAL

\int_{- \infty}^{\infty} k \sum x_{k} (t) d t = k \sum \int_{- \infty}^{\infty} x_{k} (t) d t

Always allowed for a finite number of terms, each with a finite integral. For infinitely many terms it is enough that $\sum_{k} \int ∣ x_{k} (t) ∣ d t < \infty$ , a standard result stated here without proof. Here $k$ labels the terms and each $x_{k} (t)$ is a function of $t$ .

WORKED EXAMPLE

A train of pulses is $x (t) = \sum_{k = 0}^{\infty} 2^{- k} e^{- (t - k)} u (t - k)$ volts, with $t$ in seconds. The $k$ th pulse starts at $t = k$ s with height $2^{- k}$ V. Find $\int_{- \infty}^{\infty} x (t) d t$ .

Solution

Every term is non-negative, so $\int ∣ x_{k} ∣ d t$ equals $\int x_{k} d t$ .
For the $k$ th pulse, put $τ = t - k$ . Then $\int x_{k} d t = 2^{- k} \int_{0}^{\infty} e^{- τ} d τ = 2^{- k}$ , using $a = 1$ .
The test for swapping is $\sum_{k} \int ∣ x_{k} ∣ d t = \sum_{k} 2^{- k}$ , a geometric series with $r = \frac{1}{2}$ and $∣ r ∣ < 1$ . It is finite, so the swap is allowed.
Swap and add: $\sum_{k = 0}^{\infty} 2^{- k} = \frac{1}{1 - \frac{1}{2}} = 2$ .

Answer: The area is $2$ V·s.

The sinc integral and what converges

An integral over an infinite range converges when its value settles to one finite number as each limit moves out to infinity. A transform turns $x (t)$ into a function of frequency by integrating $x (t)$ times a complex exponential. The Fourier integral uses $e^{- j ω t}$ , with $ω$ in rad/s. The Laplace integral uses $e^{- s t}$ , with $s = σ + j ω$ (Chapter 1) and $σ$ per second, so the Fourier integral is the case $σ = 0$ . As an ordinary integral, a transform exists only for the signals, and the values of $s$ , that make its integral settle.

FOURIER AND LAPLACE INTEGRALS

X (ω) = \int_{- \infty}^{\infty} x (t) e^{- j ω t} d t, X (s) = \int_{- \infty}^{\infty} x (t) e^{- s t} d t

These define the Fourier transform and the bilateral Laplace transform (integral over all $t$ ), each only where its integral converges. $X (ω)$ means the same as $X (j ω)$ . The inverse, from $X (ω)$ back to $x (t)$ , carries the factor $1/ (2 π)$ and is not needed here. Since $∣ e^{- j ω t} ∣ = 1$ , $X (ω)$ converges whenever $\int ∣ x (t) ∣ d t < \infty$ .

A finite area of $∣ x (t) ∣$ is enough for the Fourier integral to settle, but it is not needed. The normalised sinc function is $sinc (t) = sin (π t) / (π t)$ , with $sinc (0) = 1$ . The area of $∣ sinc (t) ∣$ is infinite, yet its integral, the Fourier integral at $ω = 0$ , settles. The unit step $u (t)$ has no Fourier transform as an ordinary integral; for $σ > 0$ the factor $e^{- σ t}$ makes its Laplace integral settle instead.

SINC INTEGRAL

\int_{- \infty}^{\infty} sinc (t) d t \int_{- \infty}^{\infty} \frac{sin ( a τ )}{π τ} d τ = \int_{- \infty}^{\infty} \frac{sin ( π t )}{π t} d t = 1 = 1, \int_{- \infty}^{\infty} \frac{sin ( a τ )}{τ} d τ = π (a > 0)

Stated, not derived. Each limit goes to infinity separately. The convergence is conditional: it settles only because the areas of its positive and negative lobes partly cancel, while $\int ∣ sinc (t) ∣ d t$ grows without limit. For the middle integral, put $t = a τ / π$ , so $d t = (a / π) d τ$ and $π t = a τ$ ; $a > 0$ keeps the limits in order, and $a$ cancels. Multiplying the middle integral by $π$ gives the last one. $t$ , $τ$ , $a$ are plain numbers.

DECAYING EXPONENTIAL WITH A CONVERGENCE FACTOR

\int_{- \infty}^{\infty} e^{- a t} u (t) e^{- s t} d t = \int_{0}^{\infty} e^{- (s + a) t} d t = \frac{1}{s + a}, Re (s) > - a

The set of $s$ where a Laplace integral converges is its region of convergence (ROC). For real $a$ this is the half-line integral with $a$ replaced by $s + a$ , so the ROC is $Re (s + a) > 0$ , that is $σ > - a$ . When $a \leq 0$ the Fourier integral ( $σ = 0$ ) does not converge, but every $σ > - a$ does.

The first four rows of the last column in the table below come from the half-line integral, with $a$ replaced by $s$ plus the signal's decay rate (2 for $e^{- 2 t}$ , 0 for $u (t)$ , $- 2$ for $e^{2 t}$ ). The column takes $s$ real; for complex $s$ the same ranges apply to its real part $σ$ . Only $e^{- 2∣ t ∣}$ needs a split at $t = 0$ . On $(- \infty, 0]$ the integrand is $e^{(2 - s) t}$ , which converges for $s < 2$ ; on $[0, \infty)$ it is $e^{- (2 + s) t}$ , which converges for $s > - 2$ . Both together need $- 2 < s < 2$ . The last row, $e^{t^{2}} u (t)$ , is argued by growth in its own cell.

Signal	Area of $∣ x ∣$	Fourier integral settles?	Real $s$ for which $\int x e^{- s t} d t$ converges
$e^{- 2 t} u (t)$	$1/2$	Yes	$s > - 2$
$e^{- 2∣ t ∣}$	$1$	Yes	$- 2 < s < 2$
$u (t)$	infinite	No (no ordinary Fourier transform): for $ω \neq = 0$ , $\int_{0}^{T} e^{- j ω t} d t = (1 - e^{- j ω T}) / (j ω)$ keeps circling as $T$ grows	$s > 0$
$e^{2 t} u (t)$	infinite	No: the integrand grows	$s > 2$
$e^{t^{2}} u (t)$	infinite	No: the integrand grows	None: $e^{t^{2}} e^{- s t} = e^{t (t - s)} \geq 1$ for every $t > 0$ with $t > s$ , so the integral grows without limit

WATCH OUT FOR

Thinking an integrand that tends to zero must have a finite area. Test: integrate up to $T$ and see whether the result settles. $1/ t$ tends to zero, yet $\int_{1}^{T} d t / t = ln T$ keeps growing.

Check yourself

CHECK YOURSELF 1

What is $\int_{- \infty}^{\infty} \frac{s i n ( 2 t )}{t} d t$ ?

(A)
$1$
(B)
$π$
(C)
$2 π$
(D)
It does not converge

Answer and solution

Answer: (B)

The sinc integral gives $\int \frac{s i n ( a τ )}{τ} d τ = π$ for every $a > 0$ ; here $a = 2$ .
The name of the integration variable does not change the value, so writing $t$ for $τ$ gives $\int \frac{s i n ( 2 t )}{t} d t = π$ .

CHECK YOURSELF 2

Evaluate $\int_{0}^{\infty} t e^{- 2 t} d t$ . Give the number.

Answer and solution

Answer: $0.25$ (accepted range $\pm 0.005$ )

With $a = 2 > 0$ , integration by parts gives $1/ a^{2}$ .
$1/ 2^{2} = 0.25$ .

CHECK YOURSELF 3One or more options are correct

Which of these signals have finite energy?

(A)
$e^{- t} u (t)$
(B)
$5 cos (2 t)$ for all $t$
(C)
$u (t)$
(D)
$e^{- ∣ t ∣}$

Answer and solution

Answer: (A), (D)

(A) $E = \int_{0}^{\infty} e^{- 2 t} d t = 1/2$ , finite.
(B) $5 cos (2 t)$ is periodic, and $∣ x ∣^{2} = 25 cos^{2} (2 t) = \frac{25}{2} + \frac{25}{2} cos (4 t)$ has positive area over each period, so the energy is infinite. Over a period $cos (4 t)$ averages to 0, so the power is $12.5$ , finite.
(C) $\int_{0}^{\infty} 1 d t$ diverges.
(D) $∣ x ∣^{2} = e^{- 2∣ t ∣}$ takes the same value at $t$ and $- t$ , so the area on $(- \infty, 0]$ equals the area on $[0, \infty)$ : $E = 2 \int_{0}^{\infty} e^{- 2 t} d t = 1$ , finite.

CHECK YOURSELF 4

A voltage $x (t) = 4 cos (10 t)$ V, with $t$ in seconds, is applied across 1 $Ω$ . Find its power in V $^{2}$ .

Answer and solution

Answer: $8$ (accepted range $\pm 0.01$ )

The signal is periodic with $T_{0} = 2 π /10 = π /5$ s, so $P = \frac{1}{T _{0}} \int_{0}^{T_{0}} 16 cos^{2} (10 t) d t$ .
Use $cos^{2} θ = \frac{1}{2} (1 + cos 2 θ)$ : $16 cos^{2} (10 t) = 8 + 8 cos (20 t)$ .
$cos (20 t)$ has period $2 π /20 = π /10$ s, so $T_{0} = π /5$ s holds exactly two of its periods and its integral over $T_{0}$ is 0.
So $P = \frac{1}{T _{0}} \times 8 T_{0} = 8$ .

The integrals, sums and limits that keep coming back

What you will be able to do

Before you start

Exponential integrals over half-lines

Energy, power and swapping a sum and an integral

The sinc integral and what converges

Check yourself

Used in these GATE topics