Integral cos(ax) / [(x^2+b^2)^2 dx

[laura_a]

Homework Statement

$$\int^{\infty}_{0} \frac{cosax}{(x^2+b^2)^2} dx$$ (a > 0, b>0)

the answer (from the text) is $$\frac{\pi}{4b^3}(1+ab)e^{-ab}$$

Homework Equations

I have just completed the proof for a similar question in the text (easier) which was

$$\frac{cos(ax)}{(x^2+1)} dx (a>0)$$

so I was hoping to use the proof for that in a similar way but not sure where to make changes (mainly because I don't understand Jordan's Lemma that well, even though I've read the section 100 times - I study via correspondance)

The Attempt at a Solution

The singularities are when $$(x^2+b^2)^2 =0$$
That is x = +/- ib ... (another thing I'm not sure about is how to determine which of these lies in the contour, as it is from 0 to infinity I will assume x=ib is the only singularity... is that correct?

We then have

$$\int^{R}_{0} \frac{cos(ax)}{(x^2+b^2)^2} dx = \frac{1}{2} [2\pi i \times \int_{B_{r}} \frac{exp(iaz)}{(x^2+b^2)^2}]$$

=$$\frac{1}{2} [2\pi i \times \int_{B_{R}+C_{R}} \frac{exp(iaz)}{(x^2+b^2)^2} dz - \int_{C_{R}}\frac{exp(iaz)}{(x^2+b^2)^2}]$$
Now $$\abs \int_{C_{R}} \frac{exp(iaz)}{(x^2+b^2)^2} \abs \leq \frac{1}{(R^2-b^2)^2} \times (2 \pi R) = \frac{2 \pi R}{(R^2-b^2)^2}$$

Now $$\frac{2 \pi R}{(R^2-b^2)^2}$$ goes to zero as R goes to $$\infty$$

We are then left with just the integral over the closed contour $$B_R + C_R$$

Now this is where I get stuck, because in the last question, I was able to set f(z) = 1/(z-i)(z+i) so I could easily apply the formula for residue to it, but I just don't know how to (if its possible) change the $$(x^2+b^2)^2$$ into an f(z) in terms of i...? Can anyone help me in the right direction... thanks :-)

The other question I have, is this type of proof (if it is correct) seems like a lot of writing without much actual calculation, are these types of questions similar to this? It seems once you've proven Jordans Lemma, then you can just apply the formula for residues ?

 

[mjsd]

in the case for $$f(z)=\frac{1}{(z^2+b^2)^2}$$, f(z) has a double pole at $$\pm ib$$, so rule for finding residues becomes:
For f(z) having a double pole at $$z_0$$
$$Res\left[f(z), z_0\right]=\lim_{z\rightarrow z_0}\frac{d}{dz}((z-z_0)^2f(z))$$
the proof is via Laurent expansion of f(z) about z_0

note: $$(z^2+b^2)^2=(z+ib)^2(z-ib)^2$$ can use partial fractions decomposition.. get four terms with denominator factors of
$$(z+ib),(z-ib),(z+ib)^2,(z-ib)^2$$
BUT
since you contour only enclose ONE of the two double poles, probably easier to apply the rule for finding Residue at a double pole above directly. where your f(z) would be $$f(z)=\frac{cos(ax)}{(z+ib)^2}$$ if you are considering double pole at +ib.