How to calculate induced emf of moving rod? I have the following exercise:

A copper rod with length L is moving on two frictionless conducting rails, in a distance a from an very long straight conductor. The rods resistance: R. Ignore any self inductance. The magnetic permeability is $\mu_0$
a) Find the direction and magnitude of the induced current.

I know I could probably just use faradays law finding the induced emf:  $\mathcal{E}=-\frac{d \Phi_{B}}{d t}$
But instead I tried using this other version, and I just can't get it working: $$ 
\begin{array}{l}{\mathcal{E}=\oint(\overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{B}}) \cdot d \vec{l}} \\ {\text { (all or part of a closed loop moves }} \\ {\text { in a } \overrightarrow{\boldsymbol{B}} \text { field) }}\end{array}
 $$ I think I'm a bit confused about the direction of the dL-vector
Here is my attempt:
Let say the dl-vector is along the x-axis and we integrate from a to a+L. The B-field is into the plane, and the
$$ 
\mathcal{E}=\int_{a}^{a+L}(\vec{v} \times \vec{B}(L)) \cdot d \vec{L}
 =\int_{a}^{a+L}(v \hat{k} \times B(L) \hat{\jmath}) \cdot d L \hat{i}=\int_{a}^{a+L}-v \cdot B(L) d L$$ $$=\int_{a}^{a+L}-v \cdot \frac{\mu_{0} I}{2 \pi L} d L=-\frac{v \mu_{0} I}{2 \pi}(\ln (a+L)-\ln (a)) $$ $$=\frac{v \mu_{0} I}{2 \pi} \ln \left(\frac{a}{a+L}\right)$$
I know this is the wrong emf, I'm supposed to get 
$$\mathcal{E}=\frac{v \mu_{0} I}{2 \pi} \ln \left(\frac{a+L}{L}\right)$$
And even if I chose my dL-vector to go in the other direction and integrate from a+L to a. I would get the same result.
What am I doing wrong here?

Edit:
So the reason why I thought my result was wrong, was since the answer of the induced current was:
$$ 
I=\frac{v \mu_{0} I}{2 \pi R} \ln \left(\frac{a}{a+L}\right)
 $$
But I think it comes from faradays law, where you have a negative sign in front of the induced emf. When you want to find the current you should take the magnitude. So I shouldn't multiply with the negative sign and change the fraction in the natural log. I should leave negative sign in front.
 A: 
And even if I chose my dL-vector to go in the other direction and integrate from a+L to a. I would get the same result.

This is not correct.  
Vector $d\vec L = dL\hat i$ is to be interpreted as $dL$ being the component of a step in the $\hat i$ direction.
As $dL$ is a component it can be either negative or positive.  
Once you have put limits on your integration whether the step is negative or positive is fixed and you do not need to do anything more other than evaluate the integral.  

As a simple example, Suppose you want to find the displacement when moving between position $a \hat i$ and position $b\hat i$ and the step is $d\vec x = dx\hat i$.  
Going from position $a \hat i$ to position $b\hat i$ the integral that you need to evaluate is $\displaystyle \int^{\rm b}_{\rm a} dx=(b-a)$ giving a displacement of $(b-a)\hat i$.  
If you went the other way from position $b \hat i$ to position $a\hat i$ the integral that you need to evaluate is $\displaystyle\int^{\rm a}_{\rm b} dx=(a-b)$ giving a displacement of $(a-b)\hat i$. 
Note that in both cases the indefinite integral is $\int dx$ and whether you moved in the direction of $\hat i$ or $-\hat i$ was determined by the limits of integration ie the values $a$ and $b$.  

In your example the integral which you need to evaluate is $ 
\displaystyle\int_{L_{\rm start}}^{L_{\rm finish}}-v \cdot B(L) d L$.  
Doing this when you move from $a$ to $(a+L)$ the value of the integral is $$\frac{v \mu_{0} I}{2 \pi} \ln \left(\frac{a}{a+L}\right)= - \frac{v \mu_{0} I}{2 \pi} \ln \left(\frac{a+L}{a}\right)$$ when you move from $(a+L)$ to $a$ the value of the integral is $$+ \frac{v \mu_{0} I}{2 \pi} \ln \left(\frac{a+L}{a}\right)$$ 
Think of the rod as a battery and the sign you would assign to the emf would differ if you move through the battery from the negative terminal to the positive terminal as compared with moving through the battery from the positive terminal to the negative terminal.  
The left side of the rod is the positive terminal and the right side is the negative terminal.
Outside the rod  current flows from the positive terminal to the negative terminal ie anticlockwise.  
In terms of Lenz, the magnetic flux through the loop is increasing so the induced current will try and reduce that increasing flux (produced by the $B\hat j$ field) by flowing anticlockwise producing a magnetic field in the $-\hat j$ direction.
