Why does integrating probability density yield a dimensionless number?

[Unkraut]

Hello!
I study mathematics and am in my sixth year, but...
I have a very elementary question:

I stumbled upon it while learning for quantum mechanics. But it's nothing new, it's happening to me all the time: I get confused by things like this!

Observe the following facts:
Suppose we deal with the space of wave functions over the real line.
The wave function $$\psi(x)$$ is a complex scalar. Take for example $$\psi(x)=e^{ikx}$$ (not normalizable, don't need it)
The derivative $$\psi'(x)=ike^{ikx}$$ has 1/length as it's unit.
Integrating that over some interval yields a scalar. $$\psi(b)-\psi(a)$$
The probability density $$\psi*(x)\psi(x)$$ is a scalar.
Integrating this over the real line (length) gives 1. A scalar...
Shouldn't such an operation yield a length? Am I stupid?

I am not joking. For me this is a mystery.

Thanks for your answers.

 

[inottoe]

When you integrate the probability density (scalar) with respect to length, the answer will have the same dimension (scalar x length).

 

[xts]

I can't get what you want to calculate and it seems for me that it hardly makes sense, but obvious mistake you did is that $\int_a^b\psi^{*}(x)\psi(x) dx$ is not equal 1, but $b-a$

 

[Unkraut]

I can't get what you want to calculate and it seems for me that it hardly makes sense, but obvious mistake you did is that $\int_b^a\psi^{*}(x)\psi(x) dx$ is not equal 1, but $b-a$

Sorry, I was talking about integrating over the whole (1-dimensional) space here (which has the physical dimension of length). And the total probability (of an physical wave function, not the example I used) should be 1.
But I see that my example e^ikx is not an example for a real wave function. And an actual wave function (in 1-space) actually has dimension sqrt(length), as for example the Gaussian wave packet:

$$\psi(x)=\frac{1}{\sqrt{\sqrt{2\pi}\sigma}}$$ with sigma being a length.

So my question was useless and came from wrong presumptions.