laplace transform: P(t)=sum[n=0,infty](-1)^n int[0,t]dt_{2n}

[sakaijin]

hi, i found a problem concerning the "few changes in the order of integrations" consider the function
\(\displaystyle P(t) = \sum_{n=0}^{\infty}(-1)^n\int_0^t dt_{2n}\int_0^{t_{2n}} dt_{2n-1}...\int_0^{t_2}dt_1 f(t_2-t_1)f(t_4-t_3)...f(t_{2n}-t_{2n-1})\)
now performing a Laplace transform yields
\(\displaystyle P(\lambda) = \int_0^{\infty} dt \,e^{-\lambda t} P(t)\)
Now the result after those not further specified "few changes in the order of integrations" yield.

\(\displaystyle P(\lambda) = \sum_{n=0}^{\infty}(-1)^n\int_0^{\infty} dt \int_0^{\infty} dt_1 \int_0^{\infty} dt_2...\int_0^{\infty} dt_{2n} e^{-\lambda(t_1 + t_2 + ... + t_{2n})}f(t_2)f(t_4)...f(t_{2n})\)

Is anyone able to explain me what has been done here. f(t) is a function, the form is irrelevant for this transformation it is only sufficient for the transform (smoothness, differentiability, ...).
if anyone is interested the explicit form of f(t) can also be given.
PLEASE HELP!!!

 

[vortexprof]

Re: laplace transform

i'm not sure about the details, but i would keep a couple of things in mind about these integrals:

the t[sub:66c4mqjg]i[/sub:66c4mqjg]'s are independent variables, so in the product of integrals they can be spread to show up with just their respective dt[sub:66c4mqjg]i[/sub:66c4mqjg]'s ;
the differences inside the functions will look like timeshifted arguments

hth

 

[sakaijin]

yeah but applying the following "transformation" for the arguments of the function leads to undefined integrals
\(\displaystyle t_{2n} = \tilde{t}_{2n}-\tilde{t}_{2n-1}\)
\(\displaystyle t_{2n-1} = \tilde{t}_{2n-1}\)
\(\displaystyle \Rightarrow f(t_{2n}-t_{2n-1}) = f(\tilde{t}_{2n})\)
\(\displaystyle \Rightarrow \int_0^{t_3}\,dt_2\int_0^{t_2}\,dt_1\,f(t_2-t_1) = \int_0^{\tilde{t}_3}\,(d\tilde{t}_2-d\tilde{t}_1)\underbrace{\int_0^{\tilde{t}_2-\tilde{t}_1}\,d\tilde{t}_1}_{\mbox{ill defined}}\,f(\tilde{t}_2)\)
that means a transformation of this type is NOT the solution for the problem, but the argument of the function is transformed from \(\displaystyle t_{2n}-t_{2n-1}\,\mbox{to}\,t_{2n}\).