# Complex integral limit

1. Jun 17, 2014

### mahler1

The problem statement, all variables and given/known data.

Let $\gamma_r:[0,\pi] \to \mathbb C$ be given by $\gamma_r(t)=re^{it}$. Prove that $\lim_{r \to \infty} \int_{\gamma_r} \dfrac{e^{iz}}{z}dz=0$.

The attempt at a solution.

The only thing I could do was:
$\int_{\gamma_r} \dfrac{e^{iz}}{z}dz=\int_0^{\pi} \dfrac{e^{ire^{it}}ire^{it}}{re^{it}}dt$

This last expression equals to $\int_0^{\pi} \dfrac{e^{ire^{it}}ire^{it}}{re^{it}}dt=\int_0^{\pi} e^{ire^{it}}idt$. At this point I got stuck, this expression $e^{ire^{it}}$ looks horrible to me and I don't know what to do with it.

The curve $\gamma_r$ is the upper half of the circle of radius $r$, according to the statement as the radius of the circle tends to infinity, the value of the integral goes to zero. I would like to understand intuitively why this is the case and how the function $\dfrac{e^{iz}}{z}$ behaves.

I would appreciate some help and suggestions. Thanks in advance.

2. Jun 18, 2014

### pasmith

Use the bound $$\left|\int_\gamma f(z)\,dz\right| \leq L(\gamma) \sup_\gamma |f(z)|$$ where $L(\gamma)$ is the length of the curve $\gamma$.

Note that $e^{ire^{it}} = e^{ir\cos t}e^{-r\sin t}$.

3. Jun 18, 2014

### mahler1

Thanks, that inequality is the key to the solution. Now, working on the right member of the inequality

$L(\gamma) \sup_\gamma |f(z)|=\int_0^{\pi} |rie^{it}|dt\sup_\gamma|\dfrac{e^{ir\cos(t)}e^{-r\sin(t)}}{re^{it}}|$

This last expression equals to $\int_0^{\pi} |rie^{it}|dt\sup_\gamma|\dfrac{e^{ir\cos(t)}e^{-r\sin(t)}}{re^{it}}|=r\pi \sup_{\gamma}|\dfrac{e^{ir\cos(t)}e^{-r\sin(t)}}{re^{it}}|=r\pi \sup_{\gamma}|\dfrac{e^{-r\sin(t)}}{r}|$

I have problems to find a bound for the expression $r\pi\sup_\gamma|\dfrac{e^{-r\sin(t)}}{r}|$. I would like to get to an expression with an $r$ on the denominator, something like $\dfrac{k}{r^c}$ with $k$ constant and $c\geq 1$, but I don't know how to get to that.

4. Jun 19, 2014

### dirk_mec1

The r cancels out, next the exponentional tends to zero as r tends to infinity, do you agree?

5. Jun 19, 2014

### pasmith

Unfortunately $f(z) = e^{iz}/z$ is a case where this bound doesn't give a sufficiently accurate bound for the integral. But one can instead use $$\left| \int_{t_1}^{t_2} f(Re^{it}) iRe^{it}\,dt \right| \leq R\int_{t_1}^{t_2} |f(Re^{it})|\,dt.$$ In this case you end up needing to find an estimate for $\int_0^\pi e^{-R\sin t}\,dt$. Fortunately the symmetry about $t = \pi/2$ allows the use of Jordan's Inequality to find an easily integrable $g_R: [0,\pi/2] \to \mathbb{R}$ such that $e^{-R\sin t} \leq g_R(t)$ for all $t \in [0,\pi/2]$.

6. Jun 19, 2014

### pasmith

The supremum of $e^{-r\sin t}$ on $0 \leq t \leq \pi$ (which must be calculated before taking the limit of r) is 1 (occuring at $t = 0$ and $t = \pi$), so in fact there is no exponential.

7. Jun 19, 2014

### dirk_mec1

8. Jun 19, 2014

### mahler1

Hmmm, I couldn't go further:

As you've written, it is true that $|\int_0^{\pi}f(re^{it})rie^{it}dt|\leq r\int_0^{\pi} |f(re^{it})|dt$.

The member from the right is $r\int_0^{\pi} |f(re^{it})|dt=r\int_0^{\pi} |\dfrac{e^{ir\cos(t)}e^{-r\sin(t)}}{re^{it}}|dt$.

Now, we are studying a limit for $r \to \infty$, so lets analyze this last expression for $r>1$:

Using the fact that $\dfrac{1}{r}<1$ and that $e^{-r\sin(t)}\leq 1$ (given $r>0$ and $t \in [0,\pi]$), we get

$r\int_0^{\pi} |\dfrac{e^{ir\cos(t)}e^{-r\sin(t)}}{re^{it}}|dt\leq r \int_0^{\pi}1=r\pi$

Now, arriving to that expression doesn't help at all, I would get a sort of opposite conclusion: as $r \to \infty$,$\space$ $\int_{\gamma_r}\dfrac{e^{iz}}{z}dz \to \infty$.

What else can I do to get a useful bound for the integral's absolute value?

9. Jun 19, 2014

### pasmith

This can be further simplified: both $e^{ir\cos t}$ and $e^{it}$ are equal to 1. Thus $$r \int_0^\pi |f(re^{it})|\,dt = \int_0^\pi e^{-r \sin t}\,dt.$$

You have already demonstrated by using the crude "length of curve times supremum" method that the bound $e^{-r\sin t} \leq 1$ is not good enough to do the job. You need something which actually varies with $r$.

Since $e^{-r\sin t}$ is symmetric about $t = \pi/2$ we have $$\int_0^\pi e^{-r\sin t}\,dt = 2\int_0^{\pi/2} e^{-r\sin t}\,dt.$$ If you could find an easily integrable $g_r$ such that $g_r(t) \geq e^{-r\sin t}$ for $t \in [0,\pi/2]$ and $\int_0^{\pi/2} g_r(t)\,dt \to 0$ as $r \to \infty$ then it would follow that $$\int_0^\pi e^{-r\sin t}\,dt = 2\int_0^{\pi/2} e^{-r \sin t}\,dt \leq 2\int_0^{\pi/2} g_r(t)\,dt.$$ and thus $$0 \leq \lim_{r \to \infty} \int_0^\pi e^{-r\sin t}\,dt \leq 2\lim_{r \to \infty} \int_0^{\pi/2} g_r(t)\,dt = 0.$$
Try using Jordan's Inequality as I suggested.

10. Jun 19, 2014

### mahler1

Yes, sorry, that suggestion was extremely helpful:

We want to get a bound for
$2\int_0^{\frac{\pi}{2}} e^{-r\sin(t)}dt$.

Since $r\geq 0$, the function $e^{-rx}$ is decreasing for $x\geq 0$. This means that if $x_1<x_2 \implies e^{-rx_1}\geq e^{-rx_2}$.

For $t \in [0,\dfrac{\pi}{2}]$, $\sin(t)\geq 0$, and using Jordan's inequality, we get $\dfrac{2}{\pi}t\leq \sin(t)$. But this means $e^{-r\sin(t)}\leq e^{-r\dfrac{2}{\pi}t}$.

This last inequality implies $2\int_0^{\frac{\pi}{2}} e^{-r\sin(t)}dt\leq 2\int_0^{\frac{\pi}{2}} e^{-r\frac{2}{\pi}t}dt$.

The value of that integral is $\dfrac{\pi}{r}(1-e^{-r})$ and this expression tends to zero when $r \to \infty$.

Since by a chain of inequalities, we get $0\leq |\int_{\gamma_r}\dfrac{e^{iz}}{z}dz|\leq \dfrac{\pi}{r}(1-e^{-r})$, it follows $\int_{\gamma_r}\dfrac{e^{iz}}{z}dz \to 0$ when $r \to \infty$.

pasmith, I really appreciate all the help and suggestions you gave me.