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When calculating the characteristic function of the exponential distribution function, we need to evaluate the complex-integration: \begin{align*} \int_0^\infty e^{itx}e^{-x} dx \end{align*} for any $t \in \mathbb{R}$.

I understand how to evaluate this integral by treating the real part and imaginary part separately, but I am wondering is there any approach that uses the complex analysis theory? I found some answer uses the seemingly unjustified "fundamental theorem of calculus":

$$\int_0^\infty e^{(it - 1)x} dx = \frac{1}{it - 1}\int_0^\infty e^{(it - 1)x} d(it - 1)x = \frac{1}{1 - it}.$$

I think this solution is lacking any theoretic support (maybe I am wrong, please advise if there is any theorem that supports the above calculation). Specifically, can this integral be evaluated using residue calculus (contour integration)?

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  • $\begingroup$ Thanks, but for a complex-valued function, how to justify $\int_0^\infty f'(x) dx = f(x)|_0^\infty$? Could you please provide me some references? $\endgroup$ – Zhanxiong Mar 5 '17 at 4:01
  • $\begingroup$ Integrating a complex-valued function is the same as a real-valued function. In complex analysis we look at $f(z)$ where $z$ is a complex variable, it is different because we integrate over curves in the complex plane, not over real intervals $\endgroup$ – reuns Mar 5 '17 at 4:07
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You're right to demand justification beyond glib use of the FTC. But it is true that $$ \int_0^\infty e^{-ax}dx = \frac{1}{a}$$ whenever $\Re(a)>0.$

In complex analysis, we can still affect the change of variables to $z=ax,$ but the resulting integral is along a ray in the complex plane in the direction of $a,$ not the positive real axis (unless $a$ is a positive real). We can write this (using somewhat bad notation) as $$\int_0^\infty e^{-ax}dx = \frac{1}{a} \int_0^{a\infty}e^{-z}dz.$$

(To see formally that the change of variables works, note that we have the parametrization $\gamma(t) = at$ for $0<t<\infty$ for the ray. We can write the integral of $e^{-z}/a$ along that path as $$ \int_0^\infty \frac{e^{-\gamma(t)}}{a}\gamma'(t)dt = \int_0^\infty \frac{e^{-at}}{a}adt = \int_0^\infty e^{-at}dt$$).

Now, the difficult part is how we can justify saying $$ \int_0^{a\infty}e^{-z}dz = \int_0^\infty e^{-x}dx = 1.$$

In other words we want to be able to rotate the contour back down to the real axis without changing the value of the integral.

To see why we can do this, imagine doing a integral around a large wedge-shaped contour. It goes out along the real axis to $R \gg 1$ and then goes along a circular path to $Re^{i\arg(a)}$ and then back into the origin along the ray $[0,a\infty)$ that our integral is taken over.

By Cauchy's theorem, the integral along this closed path is zero since $e^{-z}$ is analytic. The integral along the circular path goes to zero as $R\to \infty.$ We can see this cause the integrand decays like $e^{-R}.$ More formally the integral is $$ \int_0^{\arg a} e^{-Re^{i\theta}}Re^{i\theta}id\theta $$ and we have $$ \left|\int_0^{\arg a} e^{-Re^{i\theta}}Re^{i\theta}id\theta\right| \le \arg(a) \max_\theta|ie^{-Re^{i\theta}}Re^{i\theta}| = \arg(a)Re^{-R\cos(\arg(a))} \to 0$$ as $R\to\infty$

Thus as $R\to \infty,$ the integral along the real axis must cancel out the integral along the ray $[0,a\infty)$ in order that the integral along the closed path be zero as Cauchy's theorem demands. We have $$\int_0^{a\infty}e^{-z}dz = \int_0^\infty e^{-x}dx = 1 $$ and therefore $$ \int_0^\infty e^{-ax}dx = \frac{1}{a} \int_0^{a\infty}e^{-z}dz = \frac{1}{a}.$$

I've intentionally not said explicitly where $\Re(a)>0$ is used in the above argument. Of course, it's essential. See if you can find where it's used.

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  • $\begingroup$ Thanks for your answer, I found my own solution is very similar to yours! +1 $\endgroup$ – Zhanxiong Mar 5 '17 at 5:14
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I figured out a rigorous proof by myself.

If $t = 0$, then it is an integration of a real-valued function, and clearly, $\int_0^\infty e^{-x} dx = 1$.

If $t \neq 0$, without losing of generality, assume $t > 0$. Consider the contour below: enter image description here

In the picture, $n$ is a positive number that will be sent to $\infty$, the top line passes the origin and the point $(-1, t)$. And we set $f(z) = e^z, z \in \mathbb{C}$. By Cauchy's integration theorem, \begin{align} 0 = \int_\Gamma f(z) dz = \int_{\Gamma_1} e^z dz + \int_{\Gamma_2} e^z dz + \int_{\Gamma_3} e^z dz \end{align}

Let's denote the angle between $\Gamma_1$ and the real axis by $\theta_0$.

Clearly, $\int_{\Gamma_3}e^z dz = \int_{-n}^0 e^x dx = 1 - e^{-n}$.

On $\Gamma_1$, $z$ has the representation $z = (it - 1)x, x \in (0, n/|1 - it|)$, thus \begin{align*} \int_{\Gamma_1}e^z dz = \int_0^{n/\sqrt{1 + t^2}}e^{(it - 1)x}(it - 1) dx = (it - 1) \int_0^{n/\sqrt{1 + t^2}} e^{(it - 1)x} dx \end{align*}

To get the desired result, it remains to show $\int_{\Gamma_2} e^z dz \to 0$ as $n \to \infty$. Let $z = ne^{i\theta}$ with $\theta \in (\theta_0, \pi)$. It follows that \begin{align*} & \left|\int_{\Gamma_2} e^z dz\right| = \left|\int_{\theta_0}^\pi e^{ne^{i\theta}}nie^{i\theta} d\theta\right| \\ \leq & \int_{\theta_0}^\pi e^{n\cos\theta}n d\theta \\ \leq & ne^{n\cos{\theta_0}}(\pi - \theta_0) \to 0 \end{align*} as $n \to \infty$. Here we used the fact that $\pi/2 < \theta_0 < \pi$.

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The real analysis way : $$f(x) = \frac{e^{(it-1)x}}{it-1}, f'(x) = e^{(it-1)x}, \qquad \int_0^\infty f'(x)dx=f(\infty)-f(0) = \frac{1}{1-it}$$

The complex analysis way, with the (complex) change of variable $z = (it-1)u, dz = (it-1)du$ : $$\int_0^\infty e^{(it-1)u}du = \int_0^{(it-1) \infty } e^{z}\frac{dz}{it-1} = \left.\frac{e^{z}}{it-1}\right|_0^{(it-1)\infty} = \frac{1}{1-it}$$

where $ \int_0^{(it-1) \infty } $ is a contour integral

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