Use a change of contour to show that $\int_0^\infty \frac{\cos{(\alpha x)}}{x+\beta}dx = \int_0^\infty \frac{te^{-\alpha \beta t}}{t^2 + 1}dt$ A problem from an old qualifying exam:
Use a change of contour to show that
$$\int_0^\infty \frac{\cos{(\alpha x)}}{x+\beta}dx = \int_0^\infty \frac{te^{-\alpha \beta t}}{t^2 + 1}dt,$$
provided that $\alpha, \beta >0$. Define the LHS as the limit of proper integrals, and show it converges.
My attempt so far:
It seems fairly easy to tackle the last part of the question...$\cos{(\alpha x)}$ will keep picking up the same area in alternating signs, and $x$ will continue to grow, so we're basically summing up a constant times the alternating harmonic series.
I've never actually heard the phrase "change of contour". I assume what they mean is to choose a contour on one, and then use a change of variable (which will then change the contour...e.g. as $z$ traverses a certain path $z^2$ will traverse a different path). 
The right hand side looks ripe for subbing $z^2$ somehow...but then that would screw up the exponential. We need something divided by $\beta$ to get rid of the $\beta$ in the exponential on the RHS, leaving $\cos{(\alpha x)}$ as the real part of the exponential.
I also thought of trying to use a keyhole contour on the RHS and multiplying by $\log$, but it seems we'd have problems with boundedness in the left half-plane.
Any ideas or hints? I don't need you to feed me the answer. Thanks 
 A: EDIT:  I modified the contour slightly so that it is easier to show that the integral vanishes on the right side of the contour as $R \to \infty$.  I also could have used a quarter-circle in the first quadrant and Jordan's inequality.

Under the assumption that $\alpha$ and $\beta$ are positive parameters, let's integrate the function  $$ f(z) = \frac{e^{i \alpha z}}{z+\beta} $$ counterclockwise around a rectangular contour with vertices at $z=0,z=R,z=R+i\sqrt{R}$ and $z=i\sqrt{R}$, where $R$ is some large positive value.
Since there are no singularities on or inside the contour,  we get $$ \int_{0}^{R} \frac{e^{i\alpha x}}{x+\beta} \, \mathrm  dx + i \int_{0}^{\sqrt{R}} \frac{e^{i\alpha(R+it})}{(R+it)+\beta} \,  \mathrm dt - \int^{R}_{0} \frac{e^{i\alpha(t+i\sqrt{R})}}{(t+i\sqrt{R})+\beta} \, \mathrm  dt - i \int_{0}^{\sqrt{R}} \frac{e^{-\alpha t}}{it+\beta} \, \mathrm  dt =0 .  $$
As $R \to \infty$, the second integral vanishes since $$ \begin{align} \left| \int_{0}^{\sqrt{R}}  \frac{e^{i\alpha (R+it})}{(R+it)+\beta} \,  \mathrm dt \right| &\le \int_{0}^{\sqrt{R}} \frac{1}{R+\beta-t} \, \mathrm dt \quad  (\text{reverse triangle inequality}) \\ & = - \log\left(1 - \frac{\sqrt{R}}{R+\beta} \right)\to 0 \ \text{as} \ R \to \infty. \end{align}$$
And the third integral vanishes as $R \to \infty$ since
$$ \begin{align} \left|\int^{R}_{0} \frac{e^{i\alpha(t+i\sqrt{R})}}{(t+i\sqrt{R})+\beta} \, \mathrm  dt \right| &\le \int_{0}^{R} \frac{e^{-\alpha\sqrt{R}}}{\sqrt{(t+\beta)^{2}+R}} \, \mathrm  dt \\ &<\int_{0}^{R}\frac{e^{-\alpha \sqrt{R}}}{\sqrt{\beta^{2}+R}} \, \mathrm dt \\ &= \frac{Re^{-\alpha \sqrt{R}}}{\sqrt{\beta^{2}+R}}  \to 0 \ \text{as} \ R \to \infty. \end{align}$$
Therefore, we have
$$ \begin{align} \int_{0}^{\infty} \frac{e^{iax}}{x+\beta} \, \mathrm  dx &= i \int_{0}^{\infty} \frac{e^{-\alpha t}}{it+\beta} \, \mathrm  dt \\ &= i \int_{0}^{\infty} \frac{e^{-\alpha \beta u}}{i\beta u+\beta} \,  \beta \, \mathrm  du \\ &= i \int_{0}^{\infty}\frac{e^{-\alpha \beta u}}{iu+1}  \, \mathrm  du \\ &= i \int_{0}^{\infty} \frac{e^{-\alpha \beta u}(1-iu)}{u^{2}+1} \, \mathrm  du \\ &= \int_{0}^{\infty} \frac{ e^{-\alpha \beta u}(i+u)}{u^{2}+1} \, \mathrm du. \end{align}$$
And equating the real parts on both sides of the equation, we get
$$\int_{0}^{\infty} \frac{\cos (\alpha x)}{x+\beta} \, \mathrm dx = \int_{0}^{\infty} \frac{ue^{-\alpha \beta u}}{u^{2}+1} \, \mathrm  du. $$

Another less rigorous approach:
$$\begin{align} \int_{0}^{\infty} \frac{\cos (\alpha x)}{x+\beta} \, \mathrm  dx &= \int_{0}^{\infty} \int_{0}^{\infty} \cos(\alpha x) e^{-(x+\beta)t} \, \mathrm  dt \, \mathrm  dx \\ &= \int_{0}^{\infty} e^{-\beta t} \int_{0}^{\infty} \cos(\alpha x) e^{-tx} \, \mathrm  dx \, \mathrm  dt \\  &=\int_{0}^{\infty} \frac{t e^{-\beta t}}{t^{2}+\alpha ^{2}} \, \mathrm  dt \\ &= \int_{0}^{\infty} \frac{\alpha u e^{-\alpha \beta u}}{(\alpha u)^{2}+ \alpha^{2}} \,  \alpha  \, \mathrm  du \\  &= \int_{0}^{\infty} \frac{u e^{-\alpha \beta u}}{u^{2}+1} \, \mathrm  du \end{align}$$
A: Yes, it really looks like you should go for $z^2$. Let's try $x = \beta t^2$. Then we have $ dx = 2\beta tdt$ and hence:
$$ \int_0^\infty\frac{\cos(\alpha x)}{x+\beta}dx = \int_0^\infty\frac{t\cos(\alpha \beta t^2)}{t^2+1}dt $$
Maybe you should note that $\cos(x) = \frac{e^{-ix} + e^{ix}}{2}$. This would also explain the "complex analysis" tag.
