Tell me more ×
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.
up vote 2 down vote favorite

How can I evaluate this integral (without complex analysis)?

$$\int_{-\infty}^\infty\sinh [x(1-b)] \exp(iax) dx\qquad a, b\in \mathbb R$$

Thanks.

share|improve this question
1  
The way you have written this, it looks like $\sinh[a(1-b)]$ is just a constant which you can take outside of the integral. In this case the integral $\int_{-\infty}^{\infty} \exp(iax) dx$ does not converge. Maybe you meant to write something else. – Aleks Vlasev Nov 17 '11 at 9:03
Did you intend to have an $x$ somewhere in the $\sinh$? Yeah, what Aleks said ;-) – robjohn♦ Nov 17 '11 at 9:04
@AleksVlasev: Thanks, very well spotted! It's late and I'm losing it, anyway, I have edited it now. – peake Nov 17 '11 at 9:10
@robjohn: You are very right! :-) – peake Nov 17 '11 at 9:11
4  
$\sinh(x)\to\pm\infty$ as $x\to\pm\infty$ so the integral doesn't converge, no complex analysis needed :-). – robjohn♦ Nov 17 '11 at 9:12
show 1 more comment

1 Answer

up vote 1 down vote

This is the Fourier transform of $\sinh$, which can be broken down as a sum of exponentials. Because the Fourier transform is a linear operator, your integral is a sum of Fourier transforms of exponentials, that is a sum of Lorentzian functions (as it's shown for instance here).

share|improve this answer

Your Answer

 
discard

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.