Is this Definite Integral with Imaginary Bounds Real?

[Char. Limit]

So, I was playing on Wolfram Alpha, and I managed to come up with this:

http://www.wolframalpha.com/input/?i=integral_(+infinity+*+sqrt(-1)+)^pi+e^(ix)+dx&x=0&y=0

In Tex, I believe this is...

$$\int_{i\infty}^{\pi}e^{i x} dx = i$$

However, I have more than one problem with it, and I want to know if my problems are actually problems. First, the bounds. Can you multiply a transfinite number by i? Would the answer make any sense whatsoever? And can you integrate from an imaginary point to a real point?

Actually, those bounds are the only problems I have... But they do look problematic. Can someone tell me if this integral is a real integral?

 

[Dickfore]

The function $f(z) = e^{i z}$ has an absolute value:

$$|f(z)| = |e^{i z}| = e^{\Re(i z)} = e^{-\Im{z}}$$

which tends to zero as $\Im{z} \rightarrow +\infty$. The lower bound on your integral is exactly like that. Also, the function is entire. Therefore, the integral

$$F(z) = \int_{\gamma}{f(z') \, dz'}$$

has the same value for all contours $\gamma$ starting from an infinitely high point in the upper half--plane and ending anywhere in the complex plane $z$. If $z = x + i y$, it is convenient to choose the contour as:

$$\begin{array}{l} \gamma_{1}: \ z = i t, \infty > t \ge y, \ dz = i \, dt \\ \gamma_{2}: \ z = t + i y, 0 \le t \le x, \ dz = dt$$

and:

$$F(z) = \int_{\infty}^{y}{e^{i i t} \, i \, dt} + \int_{0}^{x}{e^{i (t + i y)} \, dt}$$

$$F(z) = -i \, \int_{y}^{\infty}{e^{-t} \, dt} + e^{-y} \, \int_{0}^{x}{e^{i t} \, dt}$$

$$F(z) = -i \, \left.(-e^{-t})\right|^{\infty}_{0} + e^{-y} \, \left.\frac{e^{i t}}{i}\right|^{x}_{0}$$

$$F(z) = -i + e^{-y} \frac{e^{i x} - 1}{i} = e^{-y} \, \sin x - i \, [ 1 + e^{-y} \, (\cos x - 1) ]$$