Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to calculate the following series of nested integrals with $\varepsilon(t)$ being a real function.

$$\sigma = 1 + \int_{t_0}^t\mathrm dt_1 \, \varepsilon(t_1) + \int_{t_0}^t\mathrm dt_1 \int_{t_0}^{t_1}\mathrm dt_2 \,\varepsilon(t_1)\, \varepsilon(t_2) + \int_{t_0}^t\mathrm dt_1 \int_{t_0}^{t_1}\mathrm dt_2 \int_{t_0}^{t_2}\mathrm dt_3\, \varepsilon(t_1)\, \varepsilon(t_2)\, \varepsilon(t_3) + \cdots \;.$$

The result should be

$$\sigma = \exp\left(\int_{t_0}^t\mathrm dt_1\, \varepsilon(t_1)\right) = \sum_{i=0}^\infty \frac1{i!} \left(\int_{t_0}^t\mathrm dt_1 \,\varepsilon(t_1)\right)^i \;.$$

However, comparing the series term by term I already fail to prove the equivalence for the third term. Can someone clear this up for me? Can the series be rewritten to an exponential after all? I recall from my quantum mechanics course that if $\varepsilon(t)$ was an operator, non-commuting with itself for different times, then the formal result would be

$$\sigma = T_c \exp\left(\int_{t_0}^t\mathrm dt_1\, \varepsilon(t_1)\right) \;,$$

with $T_c$ being the usual time-ordering operator. However, as I said in my case $\varepsilon(t)$ is a plain function in the real domain.

share|improve this question
    
The simplest might be to show that the right and the left hand side have the same derivative with respect to $t$. –  Fabian Apr 17 '11 at 9:34
add comment

2 Answers

up vote 7 down vote accepted

Expanding on my comment:

The function $$\sigma_1(t)= \exp\left[\int_{t_0}^t dt' \varepsilon(t')\right]$$ fulfills the following differential equation $$\sigma_1'(t)= \varepsilon(t) \sigma_1(t)$$ with the boundary condition $\sigma_1(t_0) = 1.$ We will show in a next step that $$\sigma_2 (t) = 1 + \int_{t_0}^t dt_1 \varepsilon(t_1) + \int_{t_0}^t dt_1 \int_{t_0}^{t_1} dt_2 \varepsilon(t_1) \varepsilon(t_2) + \dots \; $$ obeys the same differential equation. Because the solution to this linear differential equation is unique, it follows that $\sigma_1(t) = \sigma_2(t)$.

Taking derivative of $\sigma_2 (t)$, we recover (term by term) $$\sigma_2' (t) = \varepsilon(t) + \varepsilon(t) \int_{t_0}^t dt_1 \varepsilon(t_1) + \varepsilon(t)\int_{t_0}^t dt_1 \int_{t_0}^{t_1} dt_2 \varepsilon(t_1) \varepsilon(t_2) + \dots = \varepsilon(t)\sigma_2(t).$$ The boundary condition $\sigma_2(t_0) = 1$ also follows easily...

In conclusion, you prove with this that $$\frac1{n!} \left(\int_{t_0}^t\mathrm dt \,\varepsilon(t)\right)^n = \int_{t_0}^t\mathrm dt_1 \int_{t_0}^{t_1}\mathrm dt_2 \cdots \int_{t_0}^{t_{n-1}}\mathrm dt_n\, \varepsilon(t_1)\, \varepsilon(t_2) \cdots \varepsilon(t_n). $$

share|improve this answer
    
Wow. I'd never thought it could be done that elegant and yet that easily. Thank you so much! :) –  hennes Apr 17 '11 at 13:51
    
That is very clever. I applaud that solution. –  mixedmath Apr 21 '11 at 20:16
add comment

Rewriting your definition as $\sigma=1+I_1+I_2+I_3+\ldots$, the explanation you are after is that $(I_1)^n=(n!)I_n$ for every $n\ge1$, since then $$ \sigma=1+I_1+(I_1)^2/2!+(I_1)^3/3!+\ldots=\exp(I_1). $$ But the fact that $(I_1)^n=(n!)I_n$ is obvious: $(I_1)^n$ is the integral of the symmetric function $$ e_n:(t_1,\ldots,t_n)\mapsto\varepsilon(t_1)\cdots\varepsilon(t_n) $$ over the cube $K_n=[t_0,t]^n$. Likewise, $I_n$ is the integral of $e_n$ over the simplex $\Delta_n\subset K_n$ made of the points $(t_1,\ldots,t_n)$ such that $t_0\le t_1\le t_2\le\cdots\le t_n\le t$.

Recall that the symmetric group $\mathfrak{S}_n$ acts on $K_n$ as follows: for every $s$ in $\mathfrak{S}_n$ and $(t_1,\ldots,t_n)$ in $K_n$, $s\cdot(t_1,\ldots,t_n)=(t_{s(1)},\ldots,t_{s(n)}).$

Now, $K_n$ is the union of the $n!$ simplexes $s\cdot\Delta_n$ for $s$ in $\mathfrak{S}_n$. The function $e_n$ is symmetric hence its integral on $s\cdot\Delta_n$ is independent on $s$. The simplexes $s\cdot\Delta_n$ intersect on zero measure sets hence $(I_1)^n$ is the sum over $s$ in $\mathfrak{S}_n$ of the integrals of $e_n$ on $s\cdot\Delta_n$. Each of these integrals is equal to $I_n$ and there are $n!$ of them hence you are done.

share|improve this answer
1  
Very interesting approach, thank you! :) –  hennes Apr 17 '11 at 13:55
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.