Infinity/exponential problems.

Question

I want to evaluate $$\int_{0}^{\infty} e^{(i\alpha-1)x}\,\mathrm dx,$$ where $i$ is the imaginary number.

$$\left [ \frac{e^{(i\alpha-1)x}}{i\alpha-1}\right]_0^{\infty}$$

At this point, I beleive, if I were to substitute infinity for $x$, I would get $ e^{(i\alpha-1)\infty} $ which would just give me infinity.

However, if I take out a $-1$, I'd get:

$$\left [ \frac{e^{-1(1-i\alpha)x}}{i\alpha-1}\right]_0^{\infty}$$

which would net me $e^{-1(1-i\alpha)\infty} $ resulting in $e^{-\infty}=0$

But they are both the same value, so how would such a manipulation be getting me two different results?

Apologies for any lack of clarity or poor use of latex, it's my first time trying that.

Also, the correct result is the 2nd, where I get $0$, but I want to know what I'm missing about the first instance that gets me such a result.

Answer 1

Neither manipulation is correct!

Assuming it makes sense, the expression $e^{(i \alpha - 1) \infty}$ would be an infinite value in the direction of $i \alpha - 1$... not in the direction of the positive or negative real axis.

However, you're not supposed to substitute $\infty$ anyways; this is an improper integral; it's value is defined to be the limit

$$ \lim_{t \to +\infty} \int_0^t e^{(i \alpha - 1) x} \, dx $$

When you compute the limit (in particular, simplifying the exponential by splitting its exponent into its real and imaginary parts), it becomes clear what the correct value should be.

Answer 2

I'm assuming $\,x,\alpha\in\Bbb R\,$ , so as Hurkyl pointed out:

$$\int\limits_0^\infty e^{(i\alpha-1)x}dx=\left.\frac{1}{i\alpha-1}e^{(i\alpha-1)x}\right|_0^\infty:=\frac{1}{i\alpha-1}\lim_{b\to\infty}\left(e^{(i\alpha-1)b}-e^0\right)$$

Now, it may all depend on what values we have. For the ones with the above assumption on $\,\alpha\,$:

$$\left|\;e^{(i\alpha-1)b}\;\right|=e^{-b}\xrightarrow[b\to\infty]{}0$$

If $\,\alpha=r+is\;,\;\;r,s\in\Bbb R\,s\neq 0\,$ , then

$$\left|\;e^{(-s+ir-1)b}\;\right|=e^{-(s+1)b}\xrightarrow[b\to\infty]{}\begin{cases}0&,\;\;s+1>0\\{}\\1&,\;\;s+1=0\\{}\\\infty&,\;\;s+1<0\end{cases} $$