I guess everyone has thought about this integral and I just want to discuss what to think of the implications. So, $$\int_0^1 {\rm d}x \, (-1)^x = \int_0^1 {\rm d}x \, \left(e^{i\pi(2n+1)}\right)^x$$ where $n\in {\mathbb Z}$. Now I guess the next step implies deciding for a specific Riemann-Sheet, thus $$ \int_0^1 {\rm d}x \, e^{i\pi(2n+1)x} = \frac{2i}{\pi(2n+1)} \, . $$
So the result of the integral is multi-valued, depending on which specific Riemann-Sheet I look at.
But still one is tempted to ask, what is the intuitively correct result? The one for the principal branch, or is each result equally correct?
It is also hard to incorporate this into the picture of an integral measuring some area underneath the curve.
So does anyone have some insights or has thought about it more thoroughly?
EDIT2: Apparently it is "unclear" how to interpret a complex integral as the area under a curve. Use some imagination!
Here is an example: How would you calculate the area under a curve in a 2d case? I guess something like $$ \int_\gamma f(x,y) \, {\rm d}s $$ would qualify where ${\rm d}s = |\dot{\gamma}(t)| {\rm d}t$ is the arc-length of the curve, or? Consider $f(x,y)=x$ along the curve $\gamma:[0,1] \rightarrow {\mathbb R}^2, t \mapsto (t,t)$. Then $$ \int_\gamma x \, {\rm d}s = \sqrt{2} \int_0^1 t \, {\rm d}t = \frac{1}{\sqrt{2}} \, . $$
What would one have guessed??? The slope in $x$-direction is $1$ and in $y$-direction $0$. Therefore the directional derivative along the bisector gives $(1,0)\cdot (1/\sqrt{2},1/\sqrt{2})^t = \frac{1}{\sqrt{2}}$ and $f(s)$ is expected to be $\frac{s}{\sqrt{2}}$.
In fact the arc-length is $s=\sqrt{2}\,t$, so likewise $f(t,t)=t=\frac{s}{\sqrt{2}}=f(s)$ is a 1d function with a proper interpretation of area under it $$ \int_0^\sqrt{2} f(s) \, {\rm d}s = \frac{1}{\sqrt{2}} \int_0^\sqrt{2} s \, {\rm d}s = \frac{1}{\sqrt{2}} \, . $$
Now let's go to complex analysis. It's basically the same thing.
Let $f(z)=z=x+iy$ along the curve $\gamma:[0,1] \rightarrow {\mathbb C}, t \mapsto t+it$. The integrals over ${\rm Re}(f)$ and ${\rm Im}(f)$ can be considered in the same way the example above illustrated. So lets start with ${\rm Re}(f)$ $$ \int_\gamma {\rm Re}(f) \, {\rm d}z = \int_0^1 t \, (1+i) \, {\rm d}t = \frac{1+i}{2} $$ whose absolute value is precisely $\frac{1}{\sqrt{2}}$. Due to symmetry the result for ${\rm Im}(f)$ is the same, so adding together we have $$ \frac{1+i}{2} + i \, \frac{1+i}{2} = i $$ with absolute value $1$.
As mentioned hitherto the area of a function is interpreted as in the example from ${\mathbb R}^2$ above. ${\rm Re}(f)$ and ${\rm Im}(f)$ are thus giving areas each which have to be added vectorially.
In fact this is just what happens for 2 functions $f(x,y)=x$ and $g(x,y)=y$ in ${\mathbb R}^2$ whose areas were determined by $\frac{1}{\sqrt{2}}$. Adding vectorially leads to $$ \sqrt{\frac{1}{2} + \frac{1}{2}} = 1 $$ which is exactly the result from the complex function.
Before asking what the value of the integral is you have to decide how you are going to define $(-1)^{x}=e^{x\log (-1)}$. For each choice of $\log (-1)$ you will get one value for the integral. There is no such thing as the right value of the logarithm.