Fdtxjr

Does anybody know how to perform the following integral analytically:
beginequation
I=int^infty_eV/2 depsilon_1 int_-infty^-eV/2 depsilon_2 int^infty_0 depsilon' int_-infty^0 depsilon,delta(epsilon_2-epsilon_1+epsilon-epsilon')
endequation
Above $delta(x)$ is the Dirac's delta function.

I have used Mathematica to do this, and it gives me $I = -frac(eV)^36$.

The Dirac delta returns $1$ if its argument vanishes within the interval of integration, or zero otherwise. Doing this for the innermost integral, we get
$$
I
=int^infty_eV/2 depsilon_1 int_-infty^-eV/2 depsilon_2 int^infty_0 depsilon^prime , theta(epsilon_2-epsilon_1-epsilon^prime)
,
$$
in terms of the Heaviside theta function (i.e. $theta(x)=1$ for $x>0$, $theta(x)=0$ for $x<0$), because if $epsilon_2-epsilon_1-epsilon^prime<0$ is positive then there exists an $epsilon<0$ such that $epsilon_2-epsilon_1+epsilon-epsilon^prime=0$, and if it's negative, then $epsilon_2-epsilon_1+epsilon-epsilon^prime<0$ for all negative $epsilon$.

Pushing on, if $epsilon_2-epsilon_1$ is negative, then $epsilon_2-epsilon_1-epsilon^prime$ for all $epsilon^prime>0$, so the integral comes to zero, while if it is positive then the integral effectively runs for $epsilon^primein[0,epsilon_2-epsilon_1]$, which means that we can write
beginalign
I
& =
int^infty_eV/2 depsilon_1 int_-infty^-eV/2 depsilon_2 theta(epsilon_2-epsilon_1) int^epsilon_2-epsilon_1_0 depsilon^prime
\ & =
int^infty_eV/2 depsilon_1 int_-infty^-eV/2 depsilon_2
(epsilon_2-epsilon_1)
theta(epsilon_2-epsilon_1)
.
endalign

Furthermore, for any given $epsilon_1>eV/2$, the $epsilon_2$ integrand vanishes whenever $epsilon_2-epsilon_1<0$, i.e. when $epsilon_2<epsilon_1$, so we can get rid of the theta function and substitute in $epsilon_1$ as the lower integral limit:
beginalign
I
& =
int^infty_eV/2 depsilon_1
theta(-eV/2-epsilon_1)
int_epsilon_1^-eV/2 depsilon_2,
(epsilon_2-epsilon_1)
,
endalign
with the remaining theta function indicating that if $epsilon_1> -eV/2$ the previous theta-function integration $int_-infty^-eV/2 depsilon_2 theta(epsilon_2-epsilon_1) cdots$ has a vanishing integral throughout its domain.

Finally, to get rid of the theta functions in the integrand altogether, we note that for the integral to integrate to anything nonzero, there needs to be a range of $epsilon_1$ such that both $-eV/2-epsilon_1> 0$ and $epsilon_1>eV/2$, or in other words
$$
frac12 eV < epsilon_1 < -frac12 eV,
$$
which can only be the case if $eV<0$. To express this simply, just add one final theta function on the outside:
beginalign
I
& =
theta(-eV)
int^infty_eV/2
int_epsilon_1^-eV/2
(epsilon_2-epsilon_1)
,depsilon_2
,depsilon_1
.
endalign

The rest is just clean integration as usual.

E.P., thank you for your answer. Integrating out the first variableis the most crucial step. When I was solving this problem, I first tried to reduce the multivariable Dirac delta to a single-variable one. At the end, it became very difficult to analyze that under which circumstance does the Dirac delta become nonzero.

In the following, I start with the last equation and finish the rest of the calcuations:

beginequation
I = int^infty_eV/2depsilon_1 int^-eV/2_epsilon_1depsilon_2(epsilon_2-epsilon_1)
endequation
Here, $epsilon_1$ becomes the lower bound of $epsilon_2$, indicating that $epsilon_1 leq -eV/2$. Thus, we should change the range of the integration over $epsilon_1$ above accordingly, i.e.
beginequation
I = int^-eV/2_eV/2depsilon_1 int^-eV/2_epsilon_1depsilon_2(epsilon_2-epsilon_1)=int^-eV/2_eV/2depsilon_1 left[frac(eV)^28+fracepsilon_1^22 + fraceV2epsilon_1right]=-frac(eV)^36.quad Q.E.D.
endequation
Thanks!