Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 use the standard notations. When $x$ is real then by definition $$ I_{\nu}(x)=e^{-\nu\pi i/2}J_{\nu}(ix). $$ I want to define $I_{\nu}$ for complex $z$. Watson (Treatise of the Theory of Bessel Functions, 2nd ed, 3.7, p.77) says it is $$ \begin{cases} I_{\nu}(z)=e^{-\nu\pi i/2}J_{\nu}\left(ze^{\pi i/2}\right), & (-\pi<\arg z\leq\pi/2),\\ I_{\nu}(z)=e^{(3/2)\nu\pi i}J_{\nu}\left(ze^{-(3/2)\pi i}\right), & (\pi/2<\arg z\leq\pi). \end{cases} $$

Q1. $e^{\pi i/2}\neq e^{-(3/2)\pi i}$?

Q2. Why not good to simply put $z$ in place of $x$?

Q3. How was chosen these two cases of argument of $z$? What is the reason?

Watson says "it is usually convenient", but I want mathematical justification.

share|cite|improve this question
Should the different cases have different argument ranges? They both have $-\pi < \arg z \leq \pi/2$ now. – Antonio Vargas Oct 20 '12 at 15:30
@AntonioVargas thanks, it was a typo, I corrected it. – vesszabo Oct 23 '12 at 19:39

Your Answer


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

Browse other questions tagged or ask your own question.