2
$\begingroup$

Today I have spend a lot of time to understand one thing. I do some stupid mistake, but can not see it at all. I hope somebody open my eye to my mistake.

I a littel bit simplify the problem, to make a idea much more clear.

We have function $F=\int_0^\infty dT e^{i(m^2+ {\bf v}^2)T-\lambda T}$ where $m>0$ and $\lambda>0$ and $\bf v$ is a two dimentional vector. $\lambda$ is a small reguralization parameter. I want to calculate $\int d{\bf v} |F|^2$. First way. We can calculate firstly the $F=\frac{1}{\lambda-i(m^2+v^2)}$ After that we can set $\lambda=0$ and perform the integration over ${\bf v}$ $$ \int d{\bf v} |F|^2=\pi\int_0^\infty d v^2 \frac{1}{(m^2+v^2)^2}=\frac{\pi}{m^2} $$ It is right result. The second way $$ I=\int d{\bf v} |F|^2=\int d{\bf v} dT_1 dT_2 e^{i(m^2+ {\bf v}^2)(T_1-T_2)-\lambda (T_1+T_2)} $$ This integral is vell define. I can firstly perform the integration over ${\bf v}$ $$ I=i\pi\int \frac{dT_1 dT_2}{(T_1-T_2)} e^{i m^2(T_1-T_2)-\lambda (T_1+T_2)} $$ This integral looks like divergent in the $T_1=T_2$, but it converge becouse $I$ is a real. We do change of variables $T_1=\xi(1+u)/2$ and $T_2=\xi(1-u)/2$, we get $$ I=i\pi\int_{-1}^1\frac{ du}{2u}\int_0^\infty d\xi e^{i m^2\xi u-\lambda\xi} =-\text{Im}\,\pi\int_0^1\frac{ du}{u}\int_0^\infty d\xi e^{i m^2\xi u-\lambda\xi} $$ Than, we can integrate over $\xi$: $$ I=-\text{Im}\,\pi\int_0^1\frac{ du}{u}\frac{1}{\lambda-im^2u}=-\pi\int_0^1\frac{ du}{u}\frac{m^2 u}{\lambda^2+(m^2u)^2}=-\frac{\pi}{2}\int_{-1}^1\frac{m^2 du}{\lambda^2+(m^2u)^2}\\ =-\frac{\pi}{2\lambda}\int_{-m^2/\lambda}^{m^2/\lambda}\frac{dx}{1+x^2}=-\frac{\pi^2}{2\lambda}+\frac{\pi}{m^2}+O(\lambda) $$ We obtain the result whic contradict with the right one. The second result depends on regularization parameter $\lambda$. In the limit of $\lambda\to 0$ we obtain the $I=-\infty$. What I did wrong? How to do it correctly and reduce the first term $\sim 1/\lambda$?

In my real problem I can not do the first way because there is another functions in the exponent.

PLEASE, HELP ME. I DON'T UNDERSTAND WAT IS WRONG IN THIS CALCULATIONS!

$\endgroup$

0

You must log in to answer this question.

Browse other questions tagged .