$\int_{-\infty}^\infty e^{ikx}dx$ equals what? What would $\int\limits_{-\infty}^\infty e^{ikx}dx$ be equal to where $i$ refers to imaginary unit? What steps should I go over to solve this integral? 
I saw this in the Fourier transform, and am unsure how to solve this.
 A: Draks is right on this (+1) it makes sense as (up to a constant) a representation of the Dirac delta distribution (it is divergent from other points of view!).
More exactly :
$$2\pi \delta(k)=\int\limits_{-\infty}^\infty e^{ikx}dx$$
A: $\int\limits_{-\infty}^\infty e^{ikx}dx$ is a (EDIT: scaled by $\frac1{2\pi}$) representation of Dirac's $\delta(k)$ function. For the antiderivative see the other answers...
A: I just want to clarify what is meant by saying that "as a distribution",
$$
2\pi\delta(k) = \int_{-\infty}^\infty e^{ikx}\,dx. 
$$
The right hand side certainly doesn't make sense, and the left hand side can only make sense for $k\not=0$.  What we mean though is that for any smooth function $\phi$ with rapid decay,
$$
2\pi \phi(0) = \int_{-\infty}^\infty \int_{-\infty}^\infty e^{ikx} \phi(k) \,dk\,dx.
$$
Now both sides make sense.  This isn't trivial to prove, and in fact it implies the Fourier inversion formula (using the relationship between modulation and translation under the Fourier transform).  The proof often goes through a calculation similar to what @robjohn's answer describes.
A: The definition of the Fourier inverse:
$$ f(t) = \mathfrak{F}^{-1}\{F(jw)\} = \frac{1}{2\pi} \int_{-\infty}^{+\infty} F(jw) e^{jwt} dw $$
The Fourier pair (in the angular frequency domain):
$$ \delta(t) \leftrightarrow 1 $$
The integral in the question: 
$$ 2\pi \times \frac{1}{2\pi} \int_{-\infty}^{+\infty} 1 \times e^{jxk} dx  = 2\pi \times \delta(k) = 2\pi \delta(k)$$
The variable substitution $k=t$ was made and the u-substitution $w=x$ was made, for clarity.
A: $$
\begin{align}
\int_{-\infty}^\infty e^{ixy}\overbrace{\ \ \ e^{-\epsilon x^2}\ \ \ }^{\to1}\,\mathrm{d}x
&=e^{-\frac{y^2}{4\epsilon}}\color{#C00}{\int_{-\infty}^\infty e^{-\epsilon\left(x-\frac{iy}{2\epsilon}\right)^2}\,\mathrm{d}x}\tag{1}\\
&=\underbrace{\color{#C00}{\sqrt{\frac\pi\epsilon}}\,e^{-\frac{y^2}{4\epsilon}}}_{\to2\pi\delta(y)}\tag{2}
\end{align}
$$
Using Cauchy's Integral Theorem, the red integral in $(1)$ is simply $\int_{-\infty}^\infty e^{-\epsilon x^2}\,\mathrm{d}x=\sqrt{\frac\pi\epsilon}$ .
As $\epsilon\to0$, we get that $(2)$ approximates $2\pi\delta(y)$. That is, the integral of $(2)$ is $2\pi$ for all $\epsilon$, and as $\epsilon\to0$, the main mass of the function is squeezed into a very small region about $0$.
A: Isn't it 
$$
-{i\over k} e^{i k x} + C
$$
Just start by deriving 
$$
e^{i k x}
$$
And workout what constants are needed to adjust the result or alternatively use the formula by noticing that you are integrating
$$
e^{c x}
$$
with:
$$
c=ik
$$
A: Formally it is solved exactly as in the real case. The antiderivative is $\frac{1}{i k} e^{ikx}$. So, $\int e^{ikx} dx = \frac{1}{i k} e^{ikx} + C$.
