Complex Contour integration of rational function

[Amad27]

Hello,

Evaluate:

$$\int_{0}^{\infty} \frac{\cos(x)}{x^2 + 1} dx$$

We know that because $f(x)$ is even:$$\int_{0}^{\infty} \frac{\cos(x)}{x^2 + 1} dx = \frac{1}{2} \cdot \int_{-\infty}^{\infty} \frac{\cos(x)}{x^2 + 1} dx$$

Consider a complex function, with $z = x + iy$

$$f(z) = \frac{\cos(z)}{z^2 + 1}$$

Consider a contour $C$ on the top half of the axis, with top-semi-circle $B$, and the x-axis from $-R \to R$

We will consider:

$$\oint_{C} \frac{\cos(z)}{z^2 + 1} dz$$

Consider $z^2 + 1 = 0 \implies z = \pm i$

But because we consider the top half, only $z = i$ suffices.

$$\text{Res}_{z=i} f(z) = \lim_{z=i} (z-i) \cos(z)\cdot \frac{1}{(z-i)(z+i)}$$

$$= \frac{\cosh(1)}{2i}$$

Which implies due to the Residue theorem:

$$\oint_{C} \frac{\cos(z)}{z^2 + 1} dz = (2\pi i)\cdot \frac{\cosh(1)}{2i} = \pi\cosh(1) $$

Keeping in mind that:

$$\oint_{C} \frac{\cos(z)}{z^2 + 1} dz = \int_{-\infty}^{\infty} \frac{\cos(x)}{x^2 + 1} \,dx + \int_{B} \frac{\cos(z)}{z^2 + 1} dz = \pi\cosh(1)$$

But how do we evaluate:

$$\int_{B} \frac{\cos(z)}{z^2 + 1} dz$$

??
Thanks

 

[alyafey22]

We depend on the fact that the integral on the circle will go to zero as the diameter grows infinity. If we can't prove this then , you have to evaluate the integral on the circle , which I guess in this case will be difficult. Failing to establish that the integral vanishes on the circle suggests that the choice of the function to integrate was wrong. Actually, we avoid integrating the complex geometric functions because they are not bounded in the complex plain. So, we cannot prove that the integral vanishes on the circle. Actually we can NOT say that

$$|\cos(z)|<1 ; |\sin(z)|<1$$.

Actually , for this specific problem. We need to use the Jordan's lemma.
.

 

[Amad27]

We depend on the fact that the integral on the circle will go to zero as the diameter grows infinity. If we can't prove this then , you have to evaluate the integral on the circle , which I guess in this case will be difficult. Failing to establish that the integral vanishes on the circle suggests that the choice of the function to integrate was wrong. Actually, we avoid integrating the complex geometric functions because they are not bounded in the complex plain. So, we cannot prove that the integral vanishes on the circle. Actually we can NOT say that

$$|\cos(z)|<1 ; |\sin(z)|<1$$.

Actually , for this specific problem. We need to use the Jordan's lemma.
.

I don't know if you know about this method, but There is a method involving:

Re[$e^{iz}$] is $\cos(z)$

Because:

$e^{iz} = \cos(z) + i\sin(z)$

So the real part of that is $\cos(z)$

But then it is said that you can replace:

$$\frac{\cos(z)}{z^2 + 1} \rightarrow \frac{e^{iz}}{z^2 + 1}$$

Considering the top half of the plane.

What allows this? Because while it is true that:

Re[$e^{iz}$] = $\cos(z)$

It is not true that:

$$e^{iz} = \cos(z)$$

Idea?

 

[Euge]

Hello,

Evaluate:

$$\int_{0}^{\infty} \frac{\cos(x)}{x^2 + 1} dx$$

We know that because $f(x)$ is even:$$\int_{0}^{\infty} \frac{\cos(x)}{x^2 + 1} dx = \frac{1}{2} \cdot \int_{-\infty}^{\infty} \frac{\cos(x)}{x^2 + 1} dx$$

Consider a complex function, with $z = x + iy$

$$f(z) = \frac{\cos(z)}{z^2 + 1}$$

Consider a contour $C$ on the top half of the axis, with top-semi-circle $B$, and the x-axis from $-R \to R$

We will consider:

$$\oint_{C} \frac{\cos(z)}{z^2 + 1} dz$$

Consider $z^2 + 1 = 0 \implies z = \pm i$

But because we consider the top half, only $z = i$ suffices.

$$\text{Res}_{z=i} f(z) = \lim_{z=i} (z-i) \cos(z)\cdot \frac{1}{(z-i)(z+i)}$$

$$= \frac{\cosh(1)}{2i}$$

Which implies due to the Residue theorem:

$$\oint_{C} \frac{\cos(z)}{z^2 + 1} dz = (2\pi i)\cdot \frac{\cosh(1)}{2i} = \pi\cosh(1) $$

Keeping in mind that:

$$\oint_{C} \frac{\cos(z)}{z^2 + 1} dz = \int_{-\infty}^{\infty} \frac{\cos(x)}{x^2 + 1} \,dx + \int_{B} \frac{\cos(z)}{z^2 + 1} dz = \pi\cosh(1)$$

But how do we evaluate:

$$\int_{B} \frac{\cos(z)}{z^2 + 1} dz$$

??
Thanks

Hi Olok,

When integrating functions of the form $\cos(ax)/p(x)$ or $\sin(ax)/p(x)$, where $p$ is a real polynomial and $a \in \Bbb R$, it is generally not a good idea to integrate $\cos(az)/p(z)$ or $\sin(az)/p(z)$ over some contour. In your particular case, you'll have unboundedness issues with $\cos(z)$ on the circular boundary of $C$.

For problems like these, consider the fact that $\cos(ax) = \text{Re}(e^{iax})$ and $\sin(ax) = \text{Im}(e^{iax})$ for all $a,x\in \Bbb R$. The integrals $\int_{-\infty}^\infty \cos(ax)/p(x)\, dx$ and $\int_{-\infty}^\infty \sin(ax)/p(x)\, dx$ can be expressed as the real and imaginary parts of the integral $\int_{-\infty}^\infty e^{iax}/p(x)\, dx$, respectively. In both cases, you would consider the contour integral $\int_C e^{iaz}{p(z)}\, dz$ (where $C$ is you semicircular contour oriented counterclockwise) in order to evaluate them.

For you problem, start with the contour integral

\(\displaystyle \int_C \frac{e^{iz}}{z^2 + 1}\, dz\),

and show that the integral of $e^{iz}/(z^2 + 1)$ over the circular part of $C$ is negligible as $R \to \infty$. Technical point: Assume from the beginning that $R > 1$. It will be more helpful though if you suppose, e.g., that $R > 2$.

 

[Amad27]

Hi Olok,

When integrating functions of the form $\cos(ax)/p(x)$ or $\sin(ax)/p(x)$, where $p$ is a real polynomial and $a \in \Bbb R$, it is generally not a good idea to integrate $\cos(az)/p(z)$ or $\sin(az)/p(z)$ over some contour. In your particular case, you'll have unboundedness issues with $\cos(z)$ on the circular boundary of $C$.

For problems like these, consider the fact that $\cos(ax) = \text{Re}(e^{iax})$ and $\sin(ax) = \text{Im}(e^{iax})$ for all $a,x\in \Bbb R$. The integrals $\int_{-\infty}^\infty \cos(ax)/p(x)\, dx$ and $\int_{-\infty}^\infty \sin(ax)/p(x)\, dx$ can be expressed as the real and imaginary parts of the integral $\int_{-\infty}^\infty e^{iax}/p(x)\, dx$, respectively. In both cases, you would consider the contour integral $\int_C e^{iaz}{p(z)}\, dz$ (where $C$ is you semicircular contour oriented counterclockwise) in order to evaluate them.

For you problem, start with the contour integral

\(\displaystyle \int_C \frac{e^{iz}}{z^2 + 1}\, dz\),

and show that the integral of $e^{iz}/(z^2 + 1)$ over the circular part of $C$ is negligible as $R \to \infty$. Technical point: Assume from the beginning that $R > 1$. It will be more helpful though if you suppose, e.g., that $R > 2$.

Yes I understand the process more now.

I found from research that:

Re$\displaystyle \int_{a}^{b} f(z) dz = \displaystyle \int_{a}^{b} \text{Re}\space f(z) dz$

So we have to show that the integral is negligible. Show that:

$$\int_{C} e^{iz}/(z^2 + 1)$$

Is negligible.

So Let $R> 2$

The circumference of the top half is given by:

$$\int_{C} e^{iz}/(z^2 + 1)$$

The Radius is $R$ so circumference is also:

$P = \pi R$

But as $R$ tends to infinity $P \to \infty$ is also true. that can't happen.

So the integral doesn't tend to 0?

 

[Euge]

Olok, you neglected finding a bound for $f(z) := e^{iz}/(z^2 + 1)$ on the circular boundary. This is what I suggest. First note that on the semicircle, call it $\gamma_R$, $z = Re^{i\theta}$, where $0 \le \theta \le \pi$. So $|e^{iz}| = e^{-\text{Im}(z)} = e^{-R\sin(\theta)}$ on $\gamma_R$. Also, by the triangle inequality, $|z^2 + 1| \ge |z|^2 - 1 = R^2 - 1$ on $\gamma_R$. Thus

\(\displaystyle |f(z)| \le \frac{e^{-R\sin(\theta)}}{R^2 - 1}\)

on $\gamma_R$. Write $\gamma_R = \gamma_1 + \gamma_2$, where $\gamma_1$ is the portion of the circle on the 1st quadrant and $\gamma_2$ is the portion of the circle on the second quadrant. On $\gamma_1$, $0 \le \theta \le \pi/2$, so $\sin(\theta) \ge (2\theta)/\pi$ on $\gamma_1$. Hence \(\displaystyle \left|\int_{\gamma_1} f(z)\, dz\right| \le \int_{\gamma_1} |f(z)|\, |dz| \le \frac{1}{R^2-1}\int_0^{\pi/2} e^{-\frac{2R\theta}{\pi}}\cdot R\, d\theta = \frac{\pi(1 - e^{-R})}{2(R^2 - 1)} \to 0 \quad \text{as} \quad R \to \infty.\)

Now find a suitable estimate for

\(\displaystyle \left|\int_{\gamma_2} f(z)\, dz\right|.\)

 

[Amad27]

Olok, you neglected finding a bound for $f(z) := e^{iz}/(z^2 + 1)$ on the circular boundary. This is what I suggest. First note that on the semicircle, call it $\gamma_R$, $z = Re^{i\theta}$, where $0 \le \theta \le \pi$. So $|e^{iz}| = e^{-\text{Im}(z)} = e^{-R\sin(\theta)}$ on $\gamma_R$. Also, by the triangle inequality, $|z^2 + 1| \ge |z|^2 - 1 = R^2 - 1$ on $\gamma_R$. Thus

\(\displaystyle |f(z)| \le \frac{e^{-R\sin(\theta)}}{R^2 - 1}\)

on $\gamma_R$. Write $\gamma_R = \gamma_1 + \gamma_2$, where $\gamma_1$ is the portion of the circle on the 1st quadrant and $\gamma_2$ is the portion of the circle on the second quadrant. On $\gamma_1$, $0 \le \theta \le \pi/2$, so $\sin(\theta) \ge (2\theta)/\pi$ on $\gamma_1$. Hence \(\displaystyle \left|\int_{\gamma_1} f(z)\, dz\right| \le \int_{\gamma_1} |f(z)|\, |dz| \le \frac{1}{R^2-1}\int_0^{\pi/2} e^{-\frac{2R\theta}{\pi}}\cdot R\, d\theta = \frac{\pi(1 - e^{-R})}{2(R^2 - 1)} \to 0 \quad \text{as} \quad R \to \infty.\)

Now find a suitable estimate for

\(\displaystyle \left|\int_{\gamma_2} f(z)\, dz\right|.\)

I think I got it:

From: http://wwwf.imperial.ac.uk/~jdg/eejordan.pdf

First we must find $\lim_{R \to \infty} |f(x)|$

Okay. $R$ is on the real axis. So let us consider this.

$$\lim_{x \to \infty} f(x) = 0$$ is satisfied, and is true.

If we consider:

$F(z) = \frac{1}{z^2 + 1}$

$e^{iz}$ is seperate. our integral is in the form:

$$\oint_{C_1} \frac{e^{imz} F(z)}{1} dz$$

Because $m = 1 > 0$ the condition is satisfied.

The only singularities are the poles, those are the only places where $f(z)$ is undefined.

Therefore:

$$\lim_{R \to \infty} \oint_{C_1} \frac{e^{iz}}{z^2 + 1} dz = 0$$

The first half is:

$$\oint \frac{e^{iz}}{z^2 + 1} dz$$

Res$\displaystyle f(z, i) = \lim_{z \to i} \frac{(z-i)e^{iz}}{z^2 + 1} = \frac{e^{-1}}{2i} $

Finally,

By the residue theorem we have:

$$\oint \frac{e^{iz}}{z^2 + 1} dz = (2\pi i)\cdot \left(\frac{e^{-1}}{2i}\right) = \frac{\pi}{e} $$

Then:

$$\oint \frac{e^{iz}}{z^2 + 1} dz = \int_{-R}^{R} \frac{e^{ix}}{x^2 + 1} \,dx + \int_{C_1} f(z) dz = \frac{\pi}{e}$$

Take the limit as $R \to \infty$

$$\int_{-\infty}^{\infty} \frac{e^{ix}}{x^2 + 1} \,dx + 0 = \frac{\pi}{e}$$

Finally,

$$\int_{-\infty}^{\infty} \frac{e^{ix}}{x^2 + 1} \,dx = \frac{\pi}{e}$$

but since we replace $\cos(x)$ with $e^{ix}$ we must take the real part of the integral which is:

Re: $\frac{\pi}{e} = \frac{\pi}{e}$

$$J = \int_{-\infty}^{\infty} \frac{\cos(x)}{x^2 + 1} \,dx = \frac{\pi}{e}$$

Our original integral $I$ is half this:

$$I = \frac{J}{2}$$

$$I = \frac{\pi}{2e}$$

Which is correct.

Out of curiosity, what type of a proof were you doing for Jordan's lemma in the previous post? How did you let $z = Re^{i\theta}$?

Thanks

 

[Euge]

Hi Olok,

I didn't use Jordan's lemma. I worked from scratch. To answer your other question, on the semicircle, $z$ has modulus $R$ and argument between $0$ and $\pi$. So $z$ is of the form $Re^{i\theta}$ where $0 \le \theta \le \pi$.

 

[Amad27]

Hi Olok,

I didn't use Jordan's lemma. I worked from scratch. To answer your other question, on the semicircle, $z$ has modulus $R$ and argument between $0$ and $\pi$. So $z$ is of the form $Re^{i\theta}$ where $0 \le \theta \le \pi$.

I need to if there is a distinction between the usage of $R$ here.
-By the "modulus $R$ do you mean the radius $R$?

If you are saying:

$z = R(x + iy)$

$$z = R(\cos(\theta) + i\sin(\theta)) = Re^{i\theta}$$

$|z| = \sqrt{R^2(\cos^2(\theta) + \sin^2(\theta))} = R(1) = R$But why is it strategic to let $z = Re^{i\theta}$? Is there some motive behind this?

Also, Since you seem to be great at complex analysis, I have a harder question:

http://mathhelpboards.com/analysis-50/complex-contour-keyhole-integration-methods-13790.html#post65503

 

[Euge]

Note that even in the link you posted the author parameterized the semicircle $ H_R $ using $ z= Re^{i\theta} $. We are both using the polar representation for complex numbers to parameterize the semicircle.

 

[Amad27]

Note that even in the link you posted the author parameterized the semicircle $ H_R $ using $ z= Re^{i\theta} $. We are both using the polar representation for complex numbers to parameterize the semicircle.

I looked at your proof earlier. You stated:

$$|e^{iz}| = e^{-Im(z)} = e^{-R\sin(\theta)}$$

How did you get this?

I will try. Probably won't get it though.

$$e^{iz} = \cos(\theta) + i\sin(\theta)$$

$$|e^{iz}| = 1$$

Lets see what you meant by

$$e^{-Im(z)}$$

$$z = Re^{i\theta}$$

$$-Im(z) = -R\sin(\theta)$$

So,

$$e^{-Im(z)} = e^{-R\sin(\theta)}$$

But how does:

$$|e^{iz}| = e^{-Im(z)} = e^{-R\sin(\theta)}$$

??

$$e^{-Im(z)} \ne 1$$

I believe. So how did you get this?

 

[Euge]

In the expression $e^{iz}$, $z$ is in general complex. You're confusing it with the expression $e^{i\theta}$, where $\theta$ is a real number. The exponential function $e^z$ is defined by the equation

\(\displaystyle e^z = e^x\cos(y) + ie^x\sin(y), \quad z = x + iy \in \Bbb C\)

Since the real part of $e^z$ is $e^x\cos(y)$ and the imaginary part of $e^z$ is $e^x\sin(y)$, we have

\(\displaystyle |e^z| = \sqrt{[e^x\cos(y)]^2 + [e^x\sin(y)]^2} = e^x = e^{\text{Re}(z)}\)

The identity $|e^z| = e^{\text{Re}(z)}$ is a highly useful fact which you'll want to have in your back pocket.

Now you know that $|e^{iz}| = e^{\text{Re}(iz)}$. What is $\text{Re}(iz)$? Since $z = \text{Re}(z) + i\,\text{Im}(z)$, we have

\(\displaystyle iz = i\,\text{Re}(z) + i^2\,\text{Im}(z) = -\text{Im}(z) + i\,\text{Re}(z)\)

Therefore $\text{Re}(iz) = -\text{Im}(z)$. This is why I could write $|e^{iz}| = e^{-\text{Im}(z)}$. Given a point $z$ in the semicircle $H_R$, its polar form is $z = Re^{i\theta}$, $0 \le \theta \le \pi$. Then $z$ has imaginary part $R\sin(\theta)$, and hence $e^{-\text{Im}(z)} = e^{-R\sin(\theta)}$. This shows that, on $H_R$, $|e^{iz}| = e^{-R\sin(\theta)}$, for $0 \le \theta \le \pi$.

 

[Amad27]

In the expression $e^{iz}$, $z$ is in general complex. You're confusing it with the expression $e^{i\theta}$, where $\theta$ is a real number. The exponential function $e^z$ is defined by the equation

\(\displaystyle e^z = e^x\cos(y) + ie^x\sin(y), \quad z = x + iy \in \Bbb C\)

Since the real part of $e^z$ is $e^x\cos(y)$ and the imaginary part of $e^z$ is $e^x\sin(y)$, we have

\(\displaystyle |e^z| = \sqrt{[e^x\cos(y)]^2 + [e^x\sin(y)]^2} = e^x = e^{\text{Re}(z)}\)

The identity $|e^z| = e^{\text{Re}(z)}$ is a highly useful fact which you'll want to have in your back pocket.

Now you know that $|e^{iz}| = e^{\text{Re}(iz)}$. What is $\text{Re}(iz)$? Since $z = \text{Re}(z) + i\,\text{Im}(z)$, we have

\(\displaystyle iz = i\,\text{Re}(z) + i^2\,\text{Im}(z) = -\text{Im}(z) + i\,\text{Re}(z)\)

Therefore $\text{Re}(iz) = -\text{Im}(z)$. This is why I could write $|e^{iz}| = e^{-\text{Im}(z)}$. Given a point $z$ in the semicircle $H_R$, its polar form is $z = Re^{i\theta}$, $0 \le \theta \le \pi$. Then $z$ has imaginary part $R\sin(\theta)$, and hence $e^{-\text{Im}(z)} = e^{-R\sin(\theta)}$. This shows that, on $H_R$, $|e^{iz}| = e^{-R\sin(\theta)}$, for $0 \le \theta \le \pi$.

I thought of a new way of approaching this. We have to prove that:

$$\int_{0}^{\pi} \frac{e^{iz}}{z} dz = 0$$

Use $z = Re^{i\theta}$ to get:

$$\int_{0}^{\pi} \frac{e^{iRe^{i\theta}}i d\theta}{1}$$

$$\left| \int_{0}^{\pi} e^{iRe^{i\theta}}i d\theta \right| \le \int_{0}^{\pi} \left| e^{iRe^{i\theta}} d\theta \right|$$

I suppose I can say because,

$i^2 > i$ we have:

$$e^{iRe^{i\theta}} < e^{iRe^{i^2\theta}} = e^{Re^{-\pi}} = e^{-R}$$

$$\lim_{R \to \infty} e^{-R} = 0$$

Was this valid?

 

[Euge]

I thought of a new way of approaching this. We have to prove that:

$$\color{red}{\int_{0}^{\pi} \frac{e^{iz}}{z} dz = 0}$$

Use $z = Re^{i\theta}$ to get:

$$\int_{0}^{\pi} \frac{e^{iRe^{i\theta}}i d\theta}{1}$$

$$\left| \int_{0}^{\pi} e^{iRe^{i\theta}}i d\theta \right| \le \int_{0}^{\pi} \left| e^{iRe^{i\theta}} d\theta \right|$$

I suppose I can say because,

$\color{red}{i^2 > i}$ we have:

$$\color{red}{e^{iRe^{i\theta}} < e^{iRe^{i^2\theta}} = e^{Re^{-\pi}} = e^{-R}}$$

$$\lim_{R \to \infty} e^{-R} = 0$$

Was this valid?

No, it's not valid. The areas highlighted in red are serious errors. The first piece in red is certainly not what you're trying to prove.

 

[Amad27]

No, it's not valid. The areas highlighted in red are serious errors. The first piece in red is certainly not what you're trying to prove.

We have to prove:

$$\int_{\Gamma} e^{iz}{z(i)} dz = 0$$

$$Z = Re^{i\theta} \implies dz = iRe^{i\theta}$$

$$= \int_{0}^{\pi} \frac{e^{iRe^{i\theta}}}{1} d\theta$$

$$\left| \int_{0}^{\pi} \frac{e^{iRe^{i\theta}}}{1} d\theta \right| \le \int_{0}^{\pi} \left| e^{iRe^{i\theta}} d\theta \right|$$

If $z = x + iy$

We have $|e^z| = e^{Re(z)}$

$z = x + iy$
$iz = ix - y$

Then as you said $Re(iz) = -y = -Im(z)$

$$|e^{iz}| = e^{Re(iz)} = e^{-Im(z)}$$

Then:

$$|e^{iRe^{i\theta}}| = |e^{iz}| = e^{-Im(z)}$$

$$z = R(x + iy) = R\cos(\theta) + Ri\sin(\theta)$$

$$-Im(z) = -R\sin(\theta)$$

$$e^{-Im(z)} = e^{-R\sin(\theta)} = \frac{1}{e^{R\sin(\theta)}}$$

The perimeter is $I(l) = \pi R$

$$\lim_{R \to \infty} \frac{\pi R}{e^{R\sin(\theta)}} = 0$$

Here is the issue this requires $0 < \theta < 1$

Which is not the bound we are looking at.

What should I do?

 

[Euge]

Hi Olok,

Please look back at the post where I explained how to show that $\int_{H_R} f(z)\, dz$ is negligible as $R\to \infty$.

 

[Amad27]

Hi Olok,

Please look back at the post where I explained how to show that $\int_{H_R} f(z)\, dz$ is negligible as $R\to \infty$.

I like your approach, but I am trying to see myself how this works. I think I see it now.

We are trying to find $|e^{iz}|/|z|$

Abs = $\displaystyle \frac{e^{-R\sin(\theta)}{R} \le \frac{e^{-R\sin(\theta)}{R^2}$

Then $I(L) = \pi R$ the perimeter.

$$\lim_{R \to \infty} \frac{\pi R e^{-R\sin(\theta)}{R^2}RR

If you can give me some help on this maybe I can do it. Hopefully. Thanks =)

 

[Euge]

How does your analysis relate to the original problem? If you can explain the relationship, then perhaps I can help you better.

 

[Amad27]

How does your analysis relate to the original problem? If you can explain the relationship, then perhaps I can help you better.

I think that is the confusion, I was working on

$$\int_{0}^{\infty} \frac{\sin(x)}{x}$$

 

[Euge]

I think that is the confusion, I was working on

$$\int_{0}^{\infty} \frac{\sin(x)}{x}$$

Ok, I understand now. What was your contour?

 

[Amad27]

Ok, I understand now. What was your contour?

Hello Euge,

View attachment 3774

So then, this is the layout:

$$\oint_{C} f(z) dz = \int_{A} f(z) dz + \int_{-R}^{-\epsilon} f(x) dx + \int_{D} f(z) dz + \int_{\epsilon}^{R} f(x) dx$$

$$f(z) = \Im\frac{e^{iz}}{iz}$$

Res$_{z=0} f(z) dz = -i$

$$\oint f(z) dz = (2\pi i)(-i) = 2\pi$$

Wait, I must have done something wrong. If the other contours (circular) cancel, then this will give that the whole integral on the real axis is $2\pi$

Which is wrong?

 

[Euge]

So then, this is the layout:

$$\oint_{C} f(z) dz = \int_{A} f(z) dz + \int_{-R}^{-\epsilon} f(x) dx + \int_{D} f(z) dz + \int_{\epsilon}^{R} f(x) dx$$

$$\color{red}{f(z) = \Im\frac{e^{iz}}{iz}}$$

The formula in red is an issue. I think you meant $f(z) = \frac{e^{iz}}{z}$, in which case

$$\int_C f(z)\, dz = 2\pi i\cdot \underset{z = 0}{\operatorname{Res}} f(z) = 2\pi i \cdot 1 = 2\pi i.$$

 

[Amad27]

The formula in red is an issue. I think you meant $f(z) = \frac{e^{iz}}{z}$, in which case

$$\int_C f(z)\, dz = 2\pi i\cdot \underset{z = 0}{\operatorname{Res}} f(z) = 2\pi i \cdot 1 = 2\pi i.$$

How?

$$\Im(\sin(z)) = e^{iz}$$

Right?

So then we get:

$$f(z) = \Im \frac{e^{iz}}{z}$$

Then:

https://www.physicsforums.com/attachments/3774$$\oint_{C} f(z) dz = \int_{A} f(z) dz + \int_{-R}^{-\epsilon} f(x) dx + \int_{D} f(z) dz + \int_{\epsilon}^{R} f(x) dx = 2\pi i$$

$$\lim_{\epsilon \to 0 , R \to \infty} \oint_{C} f(z) dz = \int_{0}^{\infty} f(x) dx + \int_{-\infty}^{0} f(x) dx + \lim_{\epsilon \to 0 , R \to \infty} \int_{D} f(z) dz + \int_{A} f(z) dz = 2\pi i$$

We get:

$$\lim_{\epsilon \to 0 , R \to \infty} \oint_{C} f(z) dz = \int_{-\infty}^{\infty} f(x) dx + \lim_{\epsilon \to 0 , R \to \infty} \int_{D} f(z) dz + \int_{A} f(z) dz = 2\pi i$$

Lets see how to deal with:

$$\int_{A} f(z) dz $$

$$\left| \int_{A} f(z) dz \right| \le \int_{A} \left| \frac{e^{iz}}{z} dz \right|$$

$$|e^{iz}| = e^{-R\sin(\theta)}$$

$$\frac{|e^{iz}|}{|z|} = \frac{e^{-R\sin(\theta)}}{R}$$

By the estimation lemma,

$$\left| \int_{A} f(z) dz \right| \le (M)(L(A))$$

$L(A) = \pi R$

$$\therefore, M(L(A)) = \frac{\pi R\cdot e^{-R\sin(\theta)}}{R} = \pi\cdot e^{-R\sin(\theta)}$$

But the limit as $R \to \infty$ this will not approach $0$ what should I do?

Well, I could see one thing. Since $R \to \infty$ after all, we see that:

$$M(L(A)) = \frac{\pi R\cdot e^{-R\sin(\theta)}}{R} = \pi\cdot e^{-R\sin(\theta)} \le \pi R \cdot e^{-R\sin(\theta)}$$

Nevermind, can you guide me for this one?

 

[Euge]

Your work shows that you're doing an analysis of $\frac{e^{iz}}{z}$, not $\Im \frac{e^{iz}}{z}$.

$$\Im(\sin(z)) = e^{iz}$$

Right?

So then we get:

$$f(z) = \Im \frac{e^{iz}}{z}$$

This implication is not only incorrect, but this form of $f(z)$ does not satisfy the conditions of the Cauchy residue theorem.

Your work also shows an attempt to use the ML-inequality in order to show that the integral over the semicircle of radius $R$ is negligible as $R\to \infty$. However, as $\max_{z\in A}|f(z)| = \frac{e^R}{R}$ and $\frac{e^R}{R}$ increases without bound, the ML-inequality will not work directly. You'll need a subtler approach.

 

[Amad27]

Your work shows that you're doing an analysis of $\frac{e^{iz}}{z}$, not $\Im \frac{e^{iz}}{z}$.
This implication is not only incorrect, but this form of $f(z)$ does not satisfy the conditions of the Cauchy residue theorem.

Your work also shows an attempt to use the ML-inequality in order to show that the integral over the semicircle of radius $R$ is negligible as $R\to \infty$. However, as $\max_{z\in A}|f(z)| = \frac{e^R}{R}$ and $\frac{e^R}{R}$ increases without bound, the ML-inequality will not work directly. You'll need a subtler approach.

I almost forgot about Jordan's Lemma.

$$f(z) = \frac{e^{iz}}{z} = e^{iz}\cdot g(z)$$

Where $\displaystyle g(z) = \frac{1}{z}$

$$g(Re^{i\theta}) = \frac{1}{Re^{i\theta}} \implies \left| \frac{1}{Re^{i\theta}} \right| = \frac{1}{R} = M_R$$

$$\left| \int_{A} f(z) dz \right| \le \pi(M_R) = \frac{\pi}{R}$$

$$\lim_{R \to \infty} \frac{\pi}{R} = 0$$

So I think the best of dealing with $e^{iz}$ expressions is using Jordan's Lemma rather than the estimation lemma.

Then the other part.

$$\int_{D} f(z) dz = \int_{-\pi}^{0} (i) \cdot {e^{i\epsilon e^{i\theta}} } d\theta$$

Let $\theta \to -\theta$ then we get:

$$\int_{-\pi}^{0} (i) \cdot {e^{i\epsilon e^{i\theta}} } d\theta = (i) \int_{0}^{\pi} {e^{i\epsilon e^{i\theta}} } $$

$$\lim_{\epsilon \to 0} (i) \int_{0}^{\pi} {e^{i\epsilon e^{i\theta}} } = (i) \int_{0}^{\pi} \lim_{\epsilon \to 0} {e^{i\epsilon e^{i\theta}} } = i\pi$$

But for the dominated convergence theorem to work we need:

$$|f(z)| \le g(z)$$

$$|f(z)| = \left| {e^{i\epsilon e^{i\theta}} } \right| = e^{-\epsilon\sin(\theta)} \le e^{0} = 1$$

And

$$\int_{0}^{\pi} 1 d\theta = \pi < \infty$$

This justifies bringing the limit inside.

So then

$$\oint_{C} f(z) dz = 2\pi (i) = i\pi + \int_{-\infty}^{\infty} f(x) dx$$

$$\int_{-\infty}^{\infty} f(x) dx = i\pi$$

But because I took the linear operator $\Im e^{iz} = \sin(z)$ we will take the imaginary part of the RHS,

$$\int_{-\infty}^{\infty} f(x) dx = \Im i\pi = \pi$

This was interesting, Jordan's Lemma was the answer all along...