Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

i.e. is there a simple solution for the following integral?

$$\int_{-\infty}^{\infty} \exp(-\log^2(|\omega|/\omega_0)) \; \exp(i \omega t) \; d\omega$$

where $\omega_0 > 0$

Failing that, is there a nice series approximation?

share|cite|improve this question
Probably not. – Chris Taylor Nov 24 '11 at 8:16
@ChrisTaylor Why not? – geometrikal Nov 24 '11 at 11:36
How are you defining $\log$ when $\omega < 0$? – Willie Wong Nov 24 '11 at 11:39
@WillieWong sorry forgot to add the bars, should be absolute value of $\omega$ – geometrikal Nov 24 '11 at 12:07
up vote 2 down vote accepted

Since mention log-normal function, you probably mean to integrate $\omega$ from 0 to $\infty$, instead of using the absolute value.

The answer to your question depends on your acceptable definition of "simple" or, in other words, on the class of functions you would like to represent the integral with. Let's do some rewriting: $$ \mathcal{F}(\omega_0, t) = \omega_0 \mathcal{I}(\omega_0 t) = \omega_0 \int_0^\infty \exp(-\log^2(\omega)) \exp( i \omega \omega_0 t) \mathrm{d} \omega $$ If you insisted on integrating over the real line while using the modulus, notice that $\int_{-\infty}^\infty \exp(-\log^2 \vert \omega/\omega_0 \vert ) \exp(i \omega t) \mathrm{d} \omega = 2 \int_{0}^\infty \exp(-\log^2( \omega/\omega_0 ) ) \cos(\omega t) \mathrm{d} \omega = 2 \omega_0 \mathfrak{Re}\left( \mathcal{I}(\omega_0 t) \right)$.

The integral $\mathcal{I}(t)$ can not be represented in terms of elementary or even hypergeometric functions.

The following formal manipulations give divergent series: $$\begin{eqnarray} \mathcal{I}(t) &=& \int_0^\infty \exp(-\log^2(\omega)) \exp( i \omega t) \mathrm{d} \omega \stackrel{\omega = \exp(u)}{=} \int_{-\infty}^\infty \exp(u-u^2) \exp(i t \mathrm{e}^u ) \mathrm{d} u \\ & \stackrel{\mathrm{formal}}{=}& \sum_{k=0}^\infty \frac{(i t)^k}{k!} \int_{-\infty}^\infty \exp( (k+1) u - u^2) \mathrm{d} u = \sum_{k=0}^\infty \frac{(i t)^k}{k!} \exp\left( \frac{k^2}{4} \right) \end{eqnarray} $$ This formal series, indeed, coincides with the formal Taylor series, since $$ \mathcal{I}^{(k)}(0) = i^k \int_0^\infty \omega^k \exp(-\log^2 \omega) \mathrm{d} \omega \stackrel{\omega = \exp(u)} = i^k \exp\left( \frac{k^2}{4} \right) $$ This means that $\mathcal{I}(t)$ is not analytic at $t=0$.

Gauss-Kronrod quadratures work fine for small $t$, but Levin collocation method might be needed for large $t$. Both methods are implemented in Mathematica: enter image description here

share|cite|improve this answer
thanks for the detailed answer – geometrikal Nov 24 '11 at 21:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.