# 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

## 2 Answers

Consider $\displaystyle f(z) = \frac{e^{iaz}}{z+b}$ and integrate around a square with vertices at $z=0,z=R,z=R+iR$, and $z=iR$.

Letting $R \to \infty$, $$\int_{0}^{\infty} \frac{e^{iax}}{x+b} \ dx + i \lim_{R \to \infty} \int_{0}^{R} \frac{e^{ia(R+it})}{(R+it)+b} \ dt + \lim_{R \to \infty} \int^{0}_{R} \frac{e^{ia(t+iR)}}{(t+iR)+b} \ dt + i \int_{\infty}^{0} \frac{e^{-at}}{it+b} \ dt$$ $$=0 \$$

The second integral vanishes since

\begin{align} \Bigg|\int_{0}^{R} \frac{e^{ia(R+it})}{(R+it)+b} \ dt\Bigg| &\le \int_{0}^{R} \frac{e^{-at}}{R+b-t} \ dt \\ &= e^{-aR}e^{-ab} \int^{R+b}_{b} \frac{e^{au}}{u} \ du \\ &= e^{-aR}e^{-ab} M \to 0 \ \text{as} \ R \to \infty \end{align}

And the third integral vanishes since

\begin{align} \Bigg|\int^{R}_{0} \frac{e^{ia(t+iR)}}{(t+iR)+b} \ dt \Bigg| &\le \int_{0}^{R} \frac{e^{-aR}}{R+b-t} \ dt \\ &\le \frac{Re^{-aR}}{b} \to 0 \ \text{as} \ R \to \infty \end{align}

Therefore,

\begin{align} \int_{0}^{\infty} \frac{e^{iax}}{x+b} \ dx &= i \int_{0}^{\infty} \frac{e^{-at}}{it+b} \ dt \\ &= i \int_{0}^{\infty} \frac{e^{-abu}}{ibu+b} \ b \ du \\ &= i \int_{0}^{\infty}\frac{e^{-abu}}{1+iu} \ du \\ &= i \int_{0}^{\infty} \frac{(1-iu)e^{-abu}}{1+u^{2}} \ du \\ &= \int_{0}^{\infty} \frac{(u+i) e^{-abu}}{1+u^{2}} \ du \end{align}

And equating the real parts on both sides of the equation,

$$\int_{0}^{\infty} \frac{\cos (ax)}{x+b} = \int_{0}^{\infty} \frac{ue^{-abu}}{1+u^{2}} \ du$$

EDIT:

Another way to approach this is to note that

\begin{align} \int_{0}^{\infty} \frac{\cos (ax)}{x+b} \ dx &= \int_{0}^{\infty} \int_{0}^{\infty} \cos(ax) e^{-(x+b)t} \ dt \ dx \\ &= \int_{0}^{\infty} e^{-bt} \int_{0}^{\infty} \cos(ax) e^{-tx} \ dx \ dt \\ &=\int_{0}^{\infty} \frac{t e^{-bt}}{a^{2}+t^{2}} \ dt \\ &= \int_{0}^{\infty} \frac{au e^{-abu}}{a^{2}+(au)^{2}} a \ du \\ &= \int_{0}^{\infty} \frac{u e^{-abu}}{1+u^{2}} \ du \end{align}

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.