# Analytic Integration

1. Mar 16, 2007

### Mitra

I would like to find the attached integral analytically by using “Cauchy Residue Theorem”.
I am wondering if there is any numerical solution for this integral.

Thanks

$$\int {\frac {\exp(-M\omega) \exp(iN\omega)} {\omega^5(\frac{\omega^4}{f^4} + \frac{(\omega^2)(4\zeta^2-2)}{f^2}+ 1)}d\omega$$

from 0 to $$\infty$$

Where $$M$$, $$N$$, $$f$$ and $$\zeta$$ are known.

2. Mar 16, 2007

### Crosson

All I can do for you is find the poles and evaluate the residues. I am unable to build a complete solution that works for all values of parameters, when attempting to select a contour and see if the imaginary parts go to zero and so forth.

This is the residue of the 5th order pole at 0:

$$\frac{f^4 (M-i n)^4-24 f^2 \left(2 z^2-1\right) (M-i n)^2+24 \left(16 z^4-16 z^2+3\right)}{24 f^4}$$

Here are the four poles of the denominator which due not occur at 0:

$$\left\{-\sqrt{f^2 \left(1-2 z^2\right)-2 \sqrt{f^4 z^2 \left(z^2-1\right)}},\sqrt{f^2 \left(1-2 z^2\right)-2 \sqrt{f^4 z^2 \left(z^2-1\right)}}, -\sqrt{\left(1-2 z^2\right) f^2+2 \sqrt{f^4 z^2 \left(z^2-1\right)}},\sqrt{\left(1-2 z^2\right) f^2+2 \sqrt{f^4 z^2 \left(z^2-1\right)}}\right\}$$

Here are the four residues of the function corresponding to the four poles above:

$$\left\{\frac{e^{(M-i n) \sqrt{f^2 \left(1-2 z^2\right)-2 \sqrt{f^4 z^2 \left(z^2-1\right)}}} f^4}{8 \sqrt{f^4 z^2 \left(z^2-1\right)} \left(\left(2 z^2-1\right) f^2+2 \sqrt{f^4 z^2 \left(z^2-1\right)}\right)^3}, \frac{e^{-(M-i n) \sqrt{f^2 \left(1-2 z^2\right)-2 \sqrt{f^4 z^2 \left(z^2-1\right)}}} f^4}{8 \sqrt{f^4 z^2 \left(z^2-1\right)} \left(\left(2 z^2-1\right) f^2+2 \sqrt{f^4 z^2 \left(z^2-1\right)}\right)^3}, \frac{e^{(M-i n) \sqrt{\left(1-2 z^2\right) f^2+2 \sqrt{f^4 z^2 \left(z^2-1\right)}}} f^4}{8 \sqrt{f^4 z^2 \left(z^2-1\right)} \left(\left(1-2 z^2\right) f^2+2 \sqrt{f^4 z^2 \left(z^2-1\right)}\right)^3}, \frac{e^{-(M-i n) \sqrt{\left(1-2 z^2\right) f^2+2 \sqrt{f^4 z^2 \left(z^2-1\right)}}} f^4}{8 \sqrt{f^4 z^2 \left(z^2-1\right)} \left(\left(1-2 z^2\right) f^2+2 \sqrt{f^4 z^2 \left(z^2-1\right)}\right)^3}\right\}$$

The location of the poles depends in general on your parameters m, n , z etc., so I cannot give the final result unless you supply assumptions on the parameters, for example if M, N,f, and zeta are real and positive, we might be able to narrow it down (probably not). You mentioned integrating numerically, and that would go fine since this function is not ill-behaved, except mildy with respect to analytical methods.

Edit: Perhaps someone can help me get these LaTeX monsters to display correctly, on our precious forum of limited width.

Last edited: Mar 16, 2007
3. Mar 19, 2007

### Mitra

Residue

We assume that $$M$$, $$N$$, $$f$$and $$\zeta$$ are real and positive and $$\zeta\ll 1$$.

Actually I solved the integral but I found some discrepancies.

I would like to find the residue at following pole.
$$-f\sqrt{(1-2 \zeta^2) - i2\zeta \sqrt{ (1-\zeta^2)}}$$

As I mentioned $$\zeta\ll 1$$ so I did some simplifications and I found:

$$(\frac{1}{8 (f^4)}) (e^{-f( ( -M+N\zeta) + i( M\zeta+N)}}) (-6+\frac{i}{\zeta})$$

I would appreciate if you could help me on this residue.

Last edited: Mar 19, 2007
4. Mar 19, 2007

### Mitra

Residue of poles of 5th order at 0

The integrand has poles of 5th order at 0, to find residue may be we should

$$\lim {\frac{d^4}{d\omega^4}\frac {\omega^5 e^{-M\omega} e^{iN\omega}} {\omega^5(\frac{\omega^4}{f^4} + \frac{(\omega^2)(4\zeta^2-2)}{f^2}+ 1)}$$
$$\omega\rightarrow 0$$

and in that case the residue is zero!

Last edited: Mar 19, 2007
5. Mar 19, 2007

### Mitra

Numerical Integration

I could not obtain the numerical answer for following integrals, actually they tend to infinity.

$$\int {\frac {(e^{-M\omega}) \cos(N\omega)} {\omega^5(\frac{\omega^4}{f^4} + \frac{(\omega^2)(4\zeta^2-2)}{f^2}+ 1)}d\omega$$

$$\int {\frac {(e^{-M\omega}) \sin(N\omega)} {\omega^5(\frac{\omega^4}{f^4} + \frac{(\omega^2)(4\zeta^2-2)}{f^2}+ 1)}d\omega$$

$$\omega$$ from 0 to $$\infty$$

I am wondering if it is possible that one could be integrated analytically but not numerically.

Last edited: Mar 19, 2007