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## Homework Statement

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

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

## Homework Equations

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

[tex] \frac{cos(ax)}{(x^2+1)} dx (a>0) [/tex]

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 [tex](x^2+b^2)^2 =0[/tex]

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

[tex] \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}] [/tex]

=[tex] \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}]

[/tex]

Now [tex] \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}[/tex]

Now [tex] \frac{2 \pi R}{(R^2-b^2)^2} [/tex] goes to zero as R goes to [tex] \infty [/tex]

We are then left with just the integral over the closed contour [tex]B_R + C_R [/tex]

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 [tex] (x^2+b^2)^2[/tex] 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 with out 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 ?