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In the text "Basic Complex Analysis" Third Edition by Jerrold E. Marsden and Michael J. Hoffman I'm inquiring if there's an alternate way through Complex-Analysis to evaluate $\text{Example (4.38)}$ ? For those who don't have the textbook on hand I've written the approach the authors have taken.

$\text{Example (4.3.8)}$

Let $\omega$ be a nonzero real constant and evaluate,

$$\int_{-\infty}^{\infty} \frac{e^{i \omega x}}{1 + ix}dx.$$

$\text{Solution}$

The author notes that this is an integral of the Fourier transform type with our choice of $f$ being $1/(1+iz)$. The author notes along the real axis one can note that

$(1)$

$$|g(x)| = |e^{-i \omega z} | / | 1 + ix| = 1 / \sqrt{1 + x^{2}}.$$

It's rather trivial to see that,

$$ \int_{-\infty}^{\infty} | 1 / \sqrt{1 + x^{2}} | dx < \infty.$$

$\text{Commentary (1.0)}$

When the author noted, that the integral was of a Fourier Transform type it seems our problem boils down to evaluating the respective integrals in the form,

$(2)$

$$\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}f(z)e^{-iwz}dz$$

Or alternatively of the latter integrals of the form,

$$\int_{-\infty}^{\infty}f(z)e^{-iwz}dz $$

$\text{Remark}$

An important detail one shouldn't miss is that our choice of $f$ is seen as $f(z) = q(z)/p(z)$

To obtain convergence of our integral, the author makes note that,

$(2)$

$$\int_{-\infty}^{\infty} \frac{e^{i \omega x}}{1 + ix}dx = \lim_{A \rightarrow \infty, B \rightarrow \infty} \int_{-A}^{B}\frac{e^{-i \omega x}}{1 + ix}dx.$$

$\text{Commentary (1.1)}$

This development rather is a key ingredient since convergence is a necessary condition also one can work out that,

\begin{align*} \lim_{A \rightarrow \infty, B \rightarrow \infty} \int_{-A}^{B}\frac{e^{-i \omega x}}{1 + ix}dx&= \lim_{A \rightarrow \infty} \bigg( \int_{-A}^{B} f \, \bigg) dx \, + \lim_{B \rightarrow \infty} \bigg( \int_{-A}^{B} f \bigg)dx\\ &= \int_{-\infty}^{B} \big(f\big) \, dx + \int_{-A}^{\infty} \big(f) \, dx \end{align*}

The author notes that since $|f(z)| = 1/|1+iz| \leq 1/ (|z|-1)$ for $|z| > 1$, this factor does not shrink towards $0$. For each $\epsilon > 0$,there is an $R(\epsilon)$ such that $|f(z)| < \epsilon$ whenever $|z| \geq R(\epsilon)$. Also the author notes the exponential factor will play nice on the half plane. But which plane in question depends on the sigh of $\omega$. If $z = x+iy$ with $x$ and $y$ real, then $|e^{i \omega z}| = e^{i \omega x + \omega y} = e^{wy}$. Therefore, $\omega < 0$ implies $|e^{i \omega z} | = e^{ \omega y} \leq 1 $ in the upper half plane $\mathcal{H.}$ In a similar fashion one can say that,$\omega < 0$ implies $|e^{i \omega z} | = e^{ \omega y} \leq 1$ in the lower half plane $\mathcal{L.}$

$\text{Commentary (1.2)}$

Witnessing these ocean of developments it seems the location of the behavior depending on the sign of the exponential factor of our choice in $f$ in the planes $\mathcal{L}$ or $\mathcal{H.}$ Also as a consequence of this some of the analytic properties of our choice $f$ become restricted to either $\mathcal{L}$ or $\mathcal{H.}$ Immediate properties that come to mind are the locations of $\operatorname{Res}(f)$, where $f$ is analytic, etc. This intuition seems to be valid since the author calculates the residue's as follows

$$\operatorname{Res}\bigg(\operatorname{\frac{e^{i \omega z}}{i(z-i)}; i}\bigg) \bigg|_{z = i} = \frac{e^{\omega}}{i}.$$ The furthermore sets that,

$$\Sigma_{\mathcal{H}} = \text{the sum of the residues in } \mathcal{H}= \frac{e^{\omega}}{i}$$

$$ \Sigma_{\mathcal{L}} = \text{the sum of the residues in } \mathcal{L}=0.$$

After dealing with the location of our Residue's considers that $A$ and $B$ are both larger that $R(\epsilon)$ and larger that $1$, we can consider the rectangular paths. The rectangular contours that are considered are displayed in $\text{Figure (4.3.2)}$. He then goes on to mention that $\gamma$ is the segment of the real axis from $-A$ to $B$. We have a two closed rectangular contours denoted by $\Gamma = \gamma + \mu_{1} + \mu_{2} + \mu_{3}$ counterclockwise through the upper half plane and as a rectangle $\wedge = \gamma + \nu_{1} + \nu_{2} + \nu_{3}.$ In each case the distance $C$ the real axis will be selected larger than $R(\epsilon)$ depend on $A$ and $B$. So one will have,

$(4)$

$$ \oint_{\Gamma} g = 2\pi i \Sigma_{\mathcal{H}} = 2 \pi e^{\omega} \text{and} \oint_{\wedge} g = -2\pi i \Sigma_{\mathcal{L}}= 0. $$

Thus,

$(5)$

$$\int_{-A}^{B} \frac{e^{i \omega x}}{1 + ix}dx = \oint_{\gamma} g = 2 \pi i \Sigma_{\mathcal{H}} \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \, \, \, \, \, \, \, \, \, -\bigg( \oint_{\mu_{1}} g + \oint_{\mu_{2}} g + \oint_{\mu_{3}} g\bigg) $$

$$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = -2 \pi i \Sigma_{\mathcal{L}\quad} \! \! \! \! - \bigg( \oint_{\nu_{1}}g + \oint_{\nu_{2}}g + \oint_{\nu_{2}}g \bigg)$$ enter image description here

$$\text{Figure (4.3.2): Paths for Example (4.3.8)}$$

$\text{Commentary (1.3)}$

The author supposes that $\omega < 0$, in the case that we have favorable behavior in $\mathcal{H}\quad$ Along $\mu_{1}$, also see that $z = B + iy$, so we have that

$(6)$

\begin{align*} \lim_{\epsilon \rightarrow \infty} \bigg| \oint_{\mu_{1}}g\bigg| &= \bigg| \int_{0}^{C} f(z)e^{i \omega z}dz\bigg| \\ &= \bigg| \int_{0}^{C} f(-A+iy)e^{-i \omega(-A+iy)} idy\bigg| \leq \int_{0}^{C}\epsilon e^{\omega y}dy \leq \frac{r}{|\omega|} \rightarrow 0. \end{align*}

In order to get our desired bound over $\mu_{1}$, one has to note that,

$$\frac{e^{i \omega(B+iy)}}{1+iz} \leq \frac{i}{i}\bigg| \frac{e^{(i \omega \beta + i\omega y)}}{1 + z} \bigg| \leq \frac{i\epsilon}{i} \leq \epsilon.$$

The author notes a very similar process over $\mu_{3}$ hence,

$(7)$

\begin{align*} \lim_{\epsilon \rightarrow \infty} \bigg| \oint_{\mu_{3}}g\bigg| &= \bigg| \int_{C}^{0} f(z)e^{i \omega z}dz\bigg| \\ &= \bigg| \int_{C}^{0} f(B+iy)e^{-i \omega(B+iy)} idy\bigg| \leq \int_{C}^{0}\epsilon e^{\omega y}dy \leq \frac{\epsilon}{|\omega|} \rightarrow 0. \end{align*}

Now for the contribution over $\mu_{3}$ it's a very similar however some adjustments were made, the strict requirement is posed that $C > R(\epsilon)$ this is initially done since due to the location of $C$ it's contribution has to be small or else our estimates will be invalid. We have

$(8)$

\begin{align*} \lim_{\epsilon \rightarrow \infty} \bigg| \oint_{\mu_{3}}g\bigg| &= \bigg| \int_{B}^{-A} f(z)e^{i \omega z}dz\bigg| \\ &= \bigg| \int_{-A}^{B} f(x+iC)e^{-i \omega(x+iC)} idy\bigg| \leq \int_{C}^{0}\epsilon e^{C \omega}dy \leq \epsilon(A+B)e^{\omega C} \rightarrow 0. \end{align*}

$\text{Commentary $$(1.4)}$

The author remembers that $\omega > 0$, one can select $C$ such that $C > 1$, and also that $C > R(\epsilon)$ and large enough so that $(A + B)e^{- \omega C} < 1$. We find that for $A$ and $B$ larger then $(1)$ He discovers the bound by noting that $A$ and $B$ are respectively larger than $(1)$ and larger then $R(\epsilon)$ so finally we have,

$$\lim_{\epsilon \rightarrow \infty}\bigg| \int_{-A}^{B}g(x)dx + 2 \pi i \Sigma_{\mathcal{L}\quad} \bigg| \leq \sum_{\psi} \bigg|\oint_{\nu_{\psi}} g\bigg| \leq \bigg(\frac{2}{| \omega|} + 1 \bigg) \epsilon \rightarrow 0. $$

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1 Answer 1

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What's being done by the author here seems—to me at least—to be over-complicated. We can evaluate this integral in a much simpler manner through application of Jordan's lemma.

To evaluate

$$I(\omega)=\int_{-\infty}^\infty\frac{e^{i\omega x}}{1+ix}dx$$

we will consider a semicircular contour $\gamma_R$ of radius $R>1$ in the upper half of the complex plane like so:

                                                              Our Contour

We can think of our integral along $\gamma_R$ as the sum of two integrals

$$\oint_{\gamma_R}\frac{e^{i\omega z}}{1+iz}dz=\int_{C_R}\frac{e^{i\omega z}}{1+iz}dz+\int_{-R}^R\frac{e^{i\omega z}}{1+iz}dz\tag{1}$$

where $C_R$ is the curved portion of $\gamma_R$. If we take the limit $R\rightarrow\infty$ we see that

$$\lim_{R\rightarrow\infty}\int_{-R}^R\frac{e^{i\omega z}}{1+iz}dz=I(\omega).$$

But this is not enough for our purposes, if we are to evaluate $I(\omega)$ by the residue theorem we must also have

$$\lim_{R\rightarrow\infty}\int_{C_R}\frac{e^{i\omega z}}{1+iz}dz=0\tag{2}$$

as we want the contribution along the arc to tend to $0$.

 

To show that $(2)$ happens we may apply Jordan's lemma. We consider $\omega>0$, then applying the lemma gives

$$\begin{align*} \left | \int_{C_R}\frac{e^{i\omega z}}{1+iz}dz \right |&\leq\frac{\pi}{\omega}M_R\\ &=\frac{\pi}{\omega}\cdot\underset{\theta\in[0,\,\pi]}{\text{max}}\left | \frac{1}{1+iRe^{i\theta}}\right |\\ &=\frac{\pi}{\omega}\cdot\underset{\theta\in[0,\,\pi]}{\text{max}}\left | \frac{1-iRe^{-i\theta}}{1+R^2}\right |\\ &\leq \frac{\pi}{\omega}\cdot\frac{1+R}{1+R^2},\qquad\text{by the triangle inequality.}\tag{3} \end{align*}$$

Since the expression $(3)$ goes to $0$ for all $\omega>0$ as $R$ increases, we can back-track through our manipulations and conclude the limit in $(2)$ holds true.

 

We may now turn to the integral on the left hand side of $(1)$ and apply the residue theorem. Remembering that we specified $R>1$ we have:

$$\begin{align*} \oint_{\gamma_R}\frac{e^{i\omega z}}{1+iz}dz&=2\pi i\cdot\underset{z=i}{\text{Res}}\left[\frac{e^{i\omega z}}{1+iz}\right]\\ &=\frac{2\pi}{e^{\omega}}. \end{align*}$$

But the above holds for all $R>1$, so it holds in the limit, I.e.

$$\begin{align*} \frac{2\pi}{e^{\omega}}&=\lim_{R\rightarrow\infty}\oint_{\gamma_R}\frac{e^{i\omega z}}{1+iz}dz\\ &=\lim_{R\rightarrow\infty}\underset{\text{goes to 0}}{\underbrace{\int_{C_R}\frac{e^{i\omega z}}{1+iz}dz}}+\lim_{R\rightarrow\infty}\int_{-R}^R\frac{e^{i\omega z}}{1+iz}dz\\ &=I(\omega). \end{align*}$$

So

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

for $\omega>0.$

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  • $\begingroup$ I've gone back and reworked the answer also in the book I got this out of it seems like your proof can be generalized for a wide variety of cases. $\endgroup$
    – Zophikel
    May 30, 2019 at 20:23
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    $\begingroup$ It’s a common technique, so it generalizes very well. It’s good that you figured out whatever that book was saying haha. $\endgroup$
    – dxdydz
    May 31, 2019 at 5:48

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