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

$i$ is such that $i^2=-1$. I am not familiar with complex integral. Is $$ \frac{1}{\sqrt{2\pi t}} \int_{\mathbb{R}} e^{-\frac{(x-iut)^2}{2t}} \, dx=1 $$ as if computing the probability of a normal density function despite the mean is imaginary. .


share|cite|improve this question
up vote 6 down vote accepted

Alternate proof. Note that for complex $z$, $$ \frac{1}{\sqrt{2\pi t}} \int_{\mathbb{R}} e^{-(x-z)^2/(2t)} \, dx $$ exists, and is an entire function of $z$. But it is the constant $1$ for $z$ on the real line, so it is therefore the constant $1$ for $z$ in the whole complex plane.

share|cite|improve this answer

Yes! In fact: $\int_{-\infty}^\infty e^{-(x+b)^2/c^2}dx= c\sqrt{\pi}$. In your case $b=-iut$ and $c=\sqrt{2t}$ hence the result follows. A justification is in order: when you make the change of variables $x \to x-iut$ you just integrate on a contour (an axis) parallel to the $x$-axis and the result holds even with the presence of complex numbers in the exponential.

share|cite|improve this answer
A justification is in order. Which is missing from your post, if I am not mistaken. – Did Nov 7 '11 at 15:59

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.