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

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


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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
$\sinh(x)\to\pm\infty$ as $x\to\pm\infty$ so the integral doesn't converge, no complex analysis needed :-). – robjohn Nov 17 '11 at 9:12

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).

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