# On the value of $e^{ix}$ at $\pm \infty$

Consider the integral $$\int_{-\infty}^{+\infty} e^{ix} \, dx.$$

Integrating, we have $$\left[-ie^{ix}\vphantom{\frac11}\right]_{-\infty}^{+\infty},$$

and we need to evaluate the limits of $e^{ix}$ at ${-\infty}$ and ${+\infty}$ which, as far as I understand, do not exist since the function is oscillatory.

But if we now evaluate the integral using contour integration: $$\oint_{-\infty}^{+\infty} e^{iz} \, dz.$$ Closing in the upper half plane, so that $e^{iz} = e^{ix}e^{-y} \rightarrow 0$ as $y\rightarrow 0$, the contribution of the semi-circle to the contour goes to $0$, and we get that the integral is $0$.

What is the correct anwer and what is the reason of this inconsistency?

• On the semi-circle, not all points have large $y$. The parts of the semi-circle close to the $X$-axis still have small $y$ regardless of the radius of the semi-circle. – Pp.. Jan 2 '15 at 14:01

## 2 Answers

Matt Samuel's answer seems unclear because of what it omits. Following up on his comments: $$\int_{-R}^R e^{ix}\,dx = \left[\frac{e^{ix}}{i}\right]_{-R}^R = \frac{e^{iR}-e^{-iR}}{i} = 2\sin R$$ and that does not approach a limit as $R\to\infty$; hence the value of the improper integral does not exist.

Since the function is an entire function, its integral along a closed curve is $0$. Consequently the integral of $x\mapsto e^{ix}$ along the semicircle centered at $0$ and of radius $R$ in the upper half-plane from $+R$ to $-R$ must be $-2\sin R$, and the limit of that as $R\to\infty$ does not exist.

It is argued that for $x,y$ real, $e^{i(x+iy)}\to 0$ as $y\to+\infty$ (which is true) and that therefore (and here's the mistake) the integral along the semicircle must go to $0$.

The question is then: What is wrong with that argument? Matt Samuel's answer in its present form does not address that.

Two things:

• As the value of the function along the upper parts of the curve gets smaller, the curve gets longer. As the values go to $0$, they are multiplied by a length that goes to $\infty$, and when one quantity goes to $0$ as another goes to $\infty$, then their product may approach a finite positive number: for example $x\cdot\dfrac 1 x$ does that as $x\to+\infty$.

• The values of the function at the parts of the curve that are close to the real axis do not go to $0$ as the curve grows.

If, for example, we could say that the values of a non-negative function $f_n$ are less than $1/n$ on the interval $[0,1]$ (which interval does not get longer as $n\to\infty$, then we can say that $\displaystyle \int_0^1 f_n(x)\,dx\to0$ as $n\to\infty$. But $$\int_0^n \frac 1 n \,dx \text{ does not approach 0 as } n\to\infty$$ despite the fact that $1/n\to0$ as $n\to\infty$. And $$\int_0^1 \frac 1 {nx}\,dx\text{ does not approach 0 as }n\to\infty$$ despite the fact that $1/(nx)\to0$ as $n\to\infty$ for each value of $x$. In this second example, the problem is that the function will always be large when $x$ is near $0$.

• amazing, thanks. – SuperCiocia Jan 2 '15 at 18:05

The integral over the upper portion of the semicircle exactly cancels the integral over the real axis, so it no more has a limit than the original integral does. In fact the integral over the upper part for a semicircle of radius $r$ evaluates to exactly $-2\sin r$, cancelling the $2\sin r$ obtained by integrating over the real axis.

• But the limit as the size of the semi-circle grows is the limit as $R\to\infty$ of $0$, and certainly that limit exists. – Michael Hardy Jan 2 '15 at 14:17
• The integral of a limit is not the limit of the integral... sometimes it is, but not here. – GEdgar Jan 2 '15 at 14:18
• @GEdgar the Riemann integral is defined as the limit of the integrals over a finite path, and no limit is being taken under the integral sign. If the semicircle is not required to be centered at the origin, only have its diameter pass through it, then there is no limit. – Matt Samuel Jan 2 '15 at 14:22
• @MichaelHardy even if the semicircle is centered at the origin the limit of the upper part is not 0. We can compute it for a circle of radius $r$ to be $2\sin r$. – Matt Samuel Jan 2 '15 at 14:29
• OK, now I see what you're saying. I think you should mention that in your answer. ${}\qquad{}$ – Michael Hardy Jan 2 '15 at 14:32