2

Hi I was trying to solve this integral:

$\int_{0}^{\infty}e^{ix}dx$

When I integrate it normally and apply limits, I get an undefined answer as $e^{\infty}=\infty$

But if I take $i$ as $-\frac{1}{i}$, I get:

$\int_{0}^{\infty}e^{-\frac{1}{i}x}dx= \frac{1}{i}[e^{-\frac{1}{i}x}]_0^\infty = \frac{1}{i}[0-1]=i$

So, which one is the right answer ?

Rahul R
  • 125
  • 2

2 Answers2

3

If I am not mistaken, this integral does not converge.

\begin{align} e^{ix}=i\sin(x)+\cos(x) \end{align}

Starting here,

$$ \int_0^\infty e^{ix}\mathrm{d}x=\int_0^\infty (i\sin(x)+\cos(x))\mathrm{d}x. $$

$$ \int_0^\infty e^{ix}\mathrm{d}x=\lim_{t\rightarrow\infty}\int_0^t (i\sin(x)+\cos(x))\mathrm{d}x. $$

Both the real and imaginary limits do not exist. Therefore this integral does not converge.

I hope this helps.

PS: $\int_{-\infty}^\infty e^{ix}\mathrm{d}x$ can be computed as 0, but it still does not converge. I double checked from here.

ck1987pd
  • 1,104
  • No, $\int_{-\infty}^\infty e^{ix}\mathrm{d}x$ does not converge. That link may compute a "principal value" for the integral; but that does not mean the integral converges. – GEdgar Nov 26 '21 at 14:30
  • Ah you are right. – ck1987pd Nov 26 '21 at 14:31
  • I know that the second approach is wrong. But where did I make a mistake? What's wrong with writing i as -1/i – Rahul R Nov 26 '21 at 14:53
  • Both of them are wrong. Divergent doesn't necessarily mean infinite. We know that this integral is bounded, but still cannot be calculated. It just doesn't approach to a specific number. – ck1987pd Nov 26 '21 at 14:57
2

The integral does not converge, the problem with your second approach (and also the first one in retrospective) is the simplification at the end when writing $$ \frac{1}{i}\big[e^{-x/i}\big]_0^\infty = \frac{1}{i}[0 - 1] $$

You have to be more careful, when evaluating at infinity when complex numbers are involved. Using the definition of improper integral (to evaluate at "$\infty$") and the Euler identity to expand $e^{ix}$: $$ \lim_{x\to\infty} e^{-x/i} = \lim_{x\to\infty} e^{ix} = \lim_{x\to\infty} \big( \cos(x) + i\cdot\sin(x)\big) $$ and both the real and complex part of the limit diverges. This show that $e^{ix} \not\to \infty$ as in fact the norm of $e^{ix}$ is always 1, i.e. $|e^{ix}| = 1$.

Leo
  • 492
  • I'm still kinda confused. I know that the second approach must be wrong. But what's wrong with writing i as -1/i and applying limits? What exactly do you mean by saying that I should be more careful when complex numbers are involved? Why is it necessary to expand it using Euler identity? I need a more detailed answer. – Rahul R Nov 26 '21 at 14:57
  • 1
    There is nothing wrong with writing $i$ as $-1/i$ and applying limits. What is wrong it actually your computation of the limit itself (which is wrong in both cases). You have to be more careful because ($a>0$) $e^{a x}\to\infty$ as $x\to\infty$ only holds when $a\in\mathbb{R}$. Same with the other limit $e^{-ax}\to 0$. It is not necessary to expend it using Euler, it is just easier to see what happens this way. – Leo Nov 26 '21 at 15:03
  • So how can I compute the limits without expanding it with Euler's theorem when 'i' is written as -1/i ? You said (a>0) e^ax→∞ as x→∞ only holds when a∈R. So is there a similar formula that holds when dealing with complex integrals? – Rahul R Nov 26 '21 at 15:19
  • 3
    I can give you an intuitive way of seeing the limit. As you know $e^{i0} = 1$, $e^{i\pi}=-1$, in fact when $x$ grows, $e^{ix}$ takes all value in the complex plane which are of norm 1. $e^{ix}$ literally runs in circle as $x$ grows, this shows first that $e^{ix}$ does not go to infinity, but also that is doesn't converge to a fixed point since it goes forever in circle around 0. – Leo Nov 26 '21 at 15:25
  • Thanks. That makes sense – Rahul R Nov 26 '21 at 15:31