Why am I getting 2 different answers for Integral exp(ix) from 0 to Infinity

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 ?

Solution 1:

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. $|e^{ix}| = 1$.

Solution 2:

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.