Evalutation of loop-integrals with momentum cutoff

[salparadise]

Homework Statement

I want to calculate the integral:

$\int dk^4 \frac{1}{k^2-m^2}$

with a momentum cutoff. I know what the result should look like, but I'm not sure how to get it.

The Attempt at a Solution

I understand one needs to perform a Wick rotation, for evaluating the integral in spherical coordinates.

So: $k^0\rightarrow ik_E^0$, $k^2=-k_E^2$ and $\bar k=\sqrt{(k_E^0)^2+|\vec k|^2}$. And we obtain for the integral

$-i2\pi^2\int^\Lambda_0 d\bar k \frac{\bar k^3}{\bar k^2+m^2}$

Now, the integral is trivial to calculate,

$\int d\bar k \frac{\bar k^3}{\bar k^2+m^2}=\frac{1}{2}(\bar k^2-m^2\log(\bar k^2+m^2))$

but my problem is that I don't know how to apply the boundary values, and "revert" the Wick rotation. I assume that $\Lambda^2=|\vec k|^2$...

The result should look like
$\sim\Lambda^2+m\log\left(\frac{\Lambda^2+m^2}{m^2}\right)$

 

[NateD]

but I'm not exactly sure how to get there. Any help would be appreciated! The solution is as follows:Using the substitution \bar k=\sqrt{(k_E^0)^2+|\vec k|^2}, we can rewrite the integral as: -i2\pi^2\int^\Lambda_0 d|\vec k| \frac{|\vec k|^3}{|\vec k|^2+m^2}Now, we can perform a substitution u=|\vec k|^2 to obtain: -i2\pi^2\int^{\Lambda^2}_0 \frac{du}{2} \frac{u}{u+m^2}We can now evaluate the integral to get: -i\pi^2(\Lambda^2+m^2\log\left(\frac{\Lambda^2+m^2}{m^2}\right))Finally, we can "revert" the Wick rotation by setting k_E^0=0 and k^2=|\vec k|^2=\Lambda^2. This gives us the desired result: \sim\Lambda^2+m\log\left(\frac{\Lambda^2+m^2}{m^2}\right)