# Taking $\int_{-\infty}^{\infty} \frac{e^{i \omega x}}{1 + ix}dx$ through Residue Theory?

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,

$$(5)$$

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

$$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = -2 \pi i \Sigma_{\mathcal{L}\quad} \! \! \! \! - \bigg( \oint_{\nu_{1}}g + \oint_{\nu_{2}}g + \oint_{\nu_{2}}g \bigg)$$ $$\text{Figure (4.3.2): Paths for Example (4.3.8)}$$

$$\text{Commentary (1.3)}$$

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,

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

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:

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 $$\omega>0.$$

• 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. – Zophikel May 30 '19 at 20:23
• It’s a common technique, so it generalizes very well. It’s good that you figured out whatever that book was saying haha. – dxdydz May 31 '19 at 5:48