4
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.