Direct Integration for Continuous Characteristic Functions

Why the Support Matters in Direct Integration

A characteristic function is an expectation of a complex exponential. For a continuous random variable, that expectation becomes an integral weighted by the density. The integral is often straightforward, but its limits are essential: a density supported only on a half-line is not integrated over the whole real line, and a density with different forms on either side of a point must be integrated in pieces.

Here we calculate characteristic functions directly for uniform, exponential, and Laplace distributions. The calculations illustrate three different roles for support: finite endpoints in the uniform case, a one-sided boundary and a decaying tail in the exponential case, and two half-lines meeting at the center in the Laplace case.

Definition: If \(X\) has density \(f\) on the real line, its characteristic function is \(\varphi_X(t)=E[e^{itX}]=\int_{-\infty}^{\infty} e^{itx}f(x)\,dx\), for real \(t\), where \(i^2=-1\). Since \(|e^{itX}|=1\), this expectation exists for every real \(t\). Where \(f\) is zero, its contribution to the integral is zero.

A Useful Exponential Integral

The exponential and Laplace calculations reduce to integrals over a half-line. The following identity keeps the role of the tail explicit. If \(\alpha\) is a complex number with positive real part, then \(\int_0^\infty e^{-\alpha x}\,dx=1/\alpha\).

To justify the identity, integrate first over a finite interval. For \(R>0\), \(\int_0^R e^{-\alpha x}\,dx=(1-e^{-\alpha R})/\alpha\). Because \(|e^{-\alpha R}|=e^{-R\operatorname{Re}(\alpha)}\), this term tends to zero as \(R\) tends to infinity. The improper integral therefore converges to \(1/\alpha\). In applications below, the positive real part ensures decay even though the integrand also oscillates.

Result: For \(\lambda>0\) and real \(t\), \(\int_0^\infty e^{-(\lambda-it)x}\,dx=1/(\lambda-it)\), since \(\operatorname{Re}(\lambda-it)=\lambda>0\). The convergence condition is satisfied by the exponential decay, not by oscillation alone.

Uniform Density: Two Finite Endpoints

Let \(X\) have the uniform distribution on \([a,b]\), where \(a<b\). Its density is \(1/(b-a)\) on that interval and zero elsewhere. Integrating over the support gives the characteristic function, rather than an integral over all real numbers.

Derivation: For \(t\ne 0\), \[ \varphi_X(t)=\frac{1}{b-a}\int_a^b e^{itx}\,dx =\frac{e^{itb}-e^{ita}}{it(b-a)}. \] At \(t=0\), evaluate the defining integral directly: \(\varphi_X(0)=\int_a^b (b-a)^{-1}\,dx=1\). Thus the displayed quotient is used only for nonzero \(t\); its apparent singularity at zero is removable.

Both endpoints appear in the numerator because the antiderivative \(e^{itx}/(it)\) is evaluated at \(b\) and \(a\). The formula can also be written in a centered form. Put \(m=(a+b)/2\) and \(h=(b-a)/2\). Factoring out the contribution from the center gives, for \(t\ne0\), \(\varphi_X(t)=e^{itm}\sin(th)/(th)\). The phase \(e^{itm}\) reflects the location of the interval, while the sine ratio reflects its width.

Worked Example: A Uniform Distribution on \([0,2]\)

Let \(X\sim \operatorname{Uniform}(0,2)\). The density is \(1/2\) for \(0\le x\le2\) and zero elsewhere. For \(t\ne0\), \(\varphi_X(t)=\frac12\int_0^2 e^{itx}\,dx=(e^{2it}-1)/(2it)\). At \(t=0\), the integral of the density is \(1\), so \(\varphi_X(0)=1\).

At \(t=\pi/2\), the endpoint term is \(e^{2it}=e^{i\pi}=-1\). Hence \(\varphi_X(\pi/2)=(-1-1)/(i\pi)=-2/(i\pi)=2i/\pi\), where the last equality follows from \(1/i=-i\). Its modulus is \(2/\pi\), which is at most \(1\), as required by the modulus bound for characteristic functions.

Exponential Density: A One-Sided Support

Use the rate parameterization: an exponential random variable with rate \(\lambda>0\) has density \(\lambda e^{-\lambda x}\) for \(x\ge0\), and zero for \(x<0\). The lower support boundary at zero determines the start of the integral. The tail extends to infinity, but it is integrable because \(\lambda>0\).

Derivation: For \(X\sim\operatorname{Exponential}(\lambda)\), apply the half-line integral with \(\alpha=\lambda-it\): \[ \varphi_X(t)=\int_0^\infty e^{itx}\lambda e^{-\lambda x}\,dx =\lambda\int_0^\infty e^{-(\lambda-it)x}\,dx =\frac{\lambda}{\lambda-it}. \] The condition \(\operatorname{Re}(\lambda-it)=\lambda>0\) justifies the improper integral for every real \(t\). At zero, the formula gives \(\lambda/\lambda=1\), consistent with total probability one.

A location shift changes the phase but not the decay factor. If \(Y=c+X\), then \(e^{itY}=e^{itc}e^{itX}\), so direct substitution, or the affine transformation rule, gives \(\varphi_Y(t)=e^{itc}\lambda/(\lambda-it)\). This factor records the shift in the support boundary from zero to \(c\).

Worked Example: A Shifted Exponential with Rate 3

Let \(X=2+E\), where \(E\sim\operatorname{Exponential}(3)\). The support of \(X\) is \([2,\infty)\), and its density is \(3e^{-3(x-2)}\) there. Directly changing variables \(y=x-2\) gives \(\varphi_X(t)=\int_2^\infty e^{itx}3e^{-3(x-2)}\,dx =e^{2it}\int_0^\infty e^{ity}3e^{-3y}\,dy =e^{2it}3/(3-it)\).

At \(t=1\), rationalize the denominator by multiplying numerator and denominator by \(3+i\): \(3/(3-i)=3(3+i)/((3-i)(3+i))=3(3+i)/10=(9+3i)/10\). Therefore \(\varphi_X(1)=e^{2i}(9+3i)/10\). In particular, the coefficient is not \((3+i)/10\): the factor of \(3\) in the numerator remains after rationalization. As a check, its modulus is \(|e^{2i}|\,|9+3i|/10=\sqrt{90}/10=3/\sqrt{10}\), also equal to \(3/|3-i|\).

Laplace Density: Split at the Center

Let \(X\) have a Laplace distribution with location \(\mu\in\mathbb R\) and scale \(b>0\), using the density \(f(x)=(2b)^{-1}e^{-|x-\mu|/b}\). The absolute value in the exponent gives one expression to the left of \(\mu\) and another to its right. Splitting the integral there is necessary; treating the density as one exponential on the entire line would give an incorrect result.

Derivation: On \(x<\mu\), set \(y=\mu-x\); on \(x\ge\mu\), set \(y=x-\mu\). The two contributions are \[ \int_{-\infty}^{\mu}e^{itx}f(x)\,dx =\frac{e^{it\mu}}{2b}\int_0^\infty e^{-(1/b+it)y}\,dy =\frac{e^{it\mu}}{2(1+ibt)} \] and \[ \int_{\mu}^{\infty}e^{itx}f(x)\,dx =\frac{e^{it\mu}}{2b}\int_0^\infty e^{-(1/b-it)y}\,dy =\frac{e^{it\mu}}{2(1-ibt)}. \] Both half-line integrals converge because the real part of each exponent coefficient is \(1/b>0\). Adding the pieces yields \[ \varphi_X(t)=\frac{e^{it\mu}}{2}\left(\frac{1}{1+ibt}+\frac{1}{1-ibt}\right) =\frac{e^{it\mu}}{1+b^2t^2}. \]

The two terms are complex conjugates apart from their common phase. Their sum has no additional imaginary component: the location \(\mu\) supplies the phase, while the symmetric shape about \(\mu\) supplies the real factor \(1/(1+b^2t^2)\). At \(t=0\), the formula equals one, and direct integration also confirms that the two halves of the density each have mass \(1/2\).

Worked Example: A Laplace Distribution with Location 1 and Scale 2

Let \(X\) have the Laplace distribution with \(\mu=1\) and \(b=2\), so its density is \(\frac14e^{-|x-1|/2}\). The general formula gives \(\varphi_X(t)=e^{it}/(1+4t^2)\). At \(t=1/2\), the denominator is \(1+4(1/4)=2\), and therefore \(\varphi_X(1/2)=e^{i/2}/2=(\cos(1/2)+i\sin(1/2))/2\).

Its modulus is \(1/2\), matching \(1/(1+4t^2)\) at \(t=1/2\), since the phase \(e^{i/2}\) has modulus one. If the location were zero with the same scale, the value would instead be the real number \(1/2\); changing the location changes the phase, not this modulus.

Comparing the Support Effects

Distribution	Support used in the integral	Where the formula comes from
Uniform on \([a,b]\)	Finite interval \([a,b]\)	Evaluating an antiderivative at both endpoints
Exponential with rate \(\lambda\)	Half-line \([0,\infty)\)	A lower boundary and a decaying tail
Laplace with location \(\mu\)	Two half-lines meeting at \(\mu\)	Adding the left and right exponential contributions

These calculations matter because direct integration provides a concrete way to connect a distribution’s density to its characteristic function. The support and density shape determine which integral is evaluated; the resulting expressions can then be used in later work with characteristic functions. In practice, a reliable calculation records the support first, writes the integral over exactly that support, and checks the value at \(t=0\).

Common Pitfalls and Checks

Ignoring the support. Integrating an exponential density from \(-\infty\) to \(\infty\) is wrong when the density is supported on \([0,\infty)\). Start at the actual lower boundary.
Using one Laplace expression on both sides of the center. The absolute value changes form at \(\mu\), so split the integral there and transform each half-line separately.
Treating a formula with \(1/t\) as defined at zero. For the uniform law, compute \(\varphi_X(0)\) from the defining integral. The quotient expression is only for \(t\ne0\).
Dropping a factor while rationalizing. For example, \(3/(3-i)=3(3+i)/10\), not \((3+i)/10\). Multiply the entire numerator by the conjugate and check the resulting modulus.
Confusing an exponential rate with a scale. In this tutorial, \(\lambda\) is the rate in \(\lambda e^{-\lambda x}\), while \(b\) is the scale in the Laplace density. Keeping the density beside the formula helps prevent parameter substitutions from being reversed.

Key takeaway: A continuous characteristic function is obtained by integrating \(e^{itx}f(x)\) over the density’s support. Finite endpoints create endpoint terms, a one-sided exponential tail gives a single rational factor, and the Laplace density requires adding two half-line integrals.

Check Your Understanding

For each question, pay attention to the support and to the value at zero.

For \(X\sim\operatorname{Uniform}(-1,1)\), write the characteristic-function integral and evaluate it for \(t\ne0\). What is its value at \(t=0\)?
For an exponential random variable with rate \(4\), state the support and derive its characteristic function using the half-line integral.
How does the characteristic function of \(c+E\), where \(E\) has rate \(\lambda\), differ from that of \(E\)? Identify which factor represents the shift.
Why must the Laplace integral be split at its location parameter? State the two expressions for the density on the two sides.
For a Laplace distribution with location \(0\) and scale \(b\), is its characteristic function real or complex? Use the derived formula to justify your answer.