Another nice Laplace transform
For $alpha > -1$ and $alpha in RR$, calculate $ccL_u[t^alpha]$.
Risposte
"Eredir":
Are you sure about this result? How can a positive definite integral give a negative number?
I guess it should be $int_a^oo e^(-tau)/tau d tau = \Gamma(0, a)$. Also using Mathematica I got $\lim_{a->0^+}\Gamma(0,a) = \infty$.
In fact in my previous post I said that we can't trust the program. Moreover, Mathematica asserts that $ccL_u[1/t]=-gamma-ln(s)$.
"elgiovo":
I'm less skeptical: maybe the result $int_(0^+)^oo e^(-tau)/tau d tau=-gamma$ could turn out to be useful.
Are you sure about this result? How can a positive definite integral give a negative number?
I guess it should be $int_a^oo e^(-tau)/tau d tau = \Gamma(0, a)$. Also using Mathematica I got $\lim_{a->0^+}\Gamma(0,a) = \infty$.
"Kroldar":
I was not joking... I said that you always find a way to solve difficult problems.
Ok, I had misunderstood. Thanks for the appreciation, then!
"Kroldar":
I) Let f and g be analytical functions in a connected open. If there exists a point where f and g and all their derivatives are the same, f and g are the same everywhere.
II) Let f and g be analytical functions in a connected open. If the set of points where f and g are the same has got an accumulation point, f and g are the same everywhere.
Thank you.
"Kroldar":
I didn't know that. How did you calculate it?
Mathematica told me. Anyway I don't know if we can trust the program, since when I ask to calculate $lim_(varepsilon to 0^+)E_1(varepsilon)$ (i.e. the same integral) the answer is $oo$.
"elgiovo":
This isn't a good thing. Simple solutions are certainly preferable!
I was not joking... I said that you always find a way to solve difficult problems.
"elgiovo":
What is the statement of this II principle? And the I?
I) Let f and g be analytical functions in a connected open. If there exists a point where f and g and all their derivatives are the same, f and g are the same everywhere.
II) Let f and g be analytical functions in a connected open. If the set of points where f and g are the same has got an accumulation point, f and g are the same everywhere.
"elgiovo":
I'm less skeptical: maybe the result $int_(0^+)^oo e^(-tau)/tau d tau=-gamma$ could turn out to be useful.
I didn't know that. How did you calculate it?
"Kroldar":
Very good, your skill is always greater than the difficulties of the problems
This isn't a good thing. Simple solutions are certainly preferable!
"Kroldar":
If we remember now the II principle of identity for analytical functions, we can conclude that $ccL_u[t^(alpha)] = (Gamma(1+alpha))/(s^(1+alpha))$ for all $s$ such as $Re(s) > 0$.
What is the statement of this II principle? And the I?
"Kroldar":
I think that we can't define $ccL_u[1/t]$, as does not exist $s$ such as the function $1/t * e^(-st)$ is integrable (in some sense) around $0^+$.
I'm less skeptical: maybe the result $int_(0^+)^oo e^(-tau)/tau d tau=-gamma$ could turn out to be useful.
Very good, your skill is always greater than the difficulties of the problems 
I'll propose another solution...
You proved that $ccL_u[t^(alpha)] = (Gamma(1+alpha))/(s^(1+alpha))$ for all $s in RR^+$. If we remember now the II principle of identity for analytical functions, we can conclude that $ccL_u[t^(alpha)] = (Gamma(1+alpha))/(s^(1+alpha))$ for all $s$ such as $Re(s) > 0$.
I'll propose another solution...
"elgiovo":
$ccL_u[t^(alpha)]=int_(0^+)^(oo)t^(alpha)e^(-st)"d"t$.
Making the substitution $t to tau/s$, we obtain
$int_(0^+)^(oo)(tau/s)^(alpha)e^(-tau)("d"t)/s=1/(s^(1+alpha))int_(0^+)^(oo)tau^(alpha)e^(-tau)"d"tau$.
This is a friendly integral, in fact for $Re[alpha+1]=alpha+1>0 to alpha> -1$ it is absolutely convergent and equals $Gamma(1+alpha)$. The final result is
$ccL_u[t^(alpha)]=(Gamma(1+alpha))/(s^(1+alpha))$.
You proved that $ccL_u[t^(alpha)] = (Gamma(1+alpha))/(s^(1+alpha))$ for all $s in RR^+$. If we remember now the II principle of identity for analytical functions, we can conclude that $ccL_u[t^(alpha)] = (Gamma(1+alpha))/(s^(1+alpha))$ for all $s$ such as $Re(s) > 0$.
Applying the estimation lemma, we see that $|int_("arc")z^(alpha)e^(-z)"d"z|<= phi*R*max_(z in "arc")|z^alpha e^(-z)|$. We have $|z^alpha e^(-z)|=R^alpha e^(-R cos (xi))$, with $-phi>0$ the integral along the arc vanishes.
"Kroldar":
When $t to 0$ also $tau to 0$, but when $t to +oo$ then $tau$ is not real anymore, so we can't do an integration along a real semiaxis.
Yes, of course. So we have to integrate along a line forming an angle $-pi/2
The line integral would then evaluate to $oint = int_([0,Re^(i phi)]) + int_("arc") + int_([R,0])=0$, or $int_([0,Re^(i phi)])=oint - int_("arc") -int_([R,0])=int_([0,R])- int_("arc")$, so that if $int_("arc")$ was kind enough to disappear with $R to oo$, the result would be the same as in the real case. Am I on the wrong track?
When $t to 0$ also $tau to 0$, but when $t to +oo$ then $tau$ is not real anymore, so we can't do an integration along a real semiaxis.
I think that we can't define $ccL_u[1/t]$, as does not exist $s$ such as the function $1/t * e^(-st)$ is integrable (in some sense) around $0^+$.
"elgiovo":
BONUS QUESTION (I still don't know the answer): what if $alpha=-1$?
I think that we can't define $ccL_u[1/t]$, as does not exist $s$ such as the function $1/t * e^(-st)$ is integrable (in some sense) around $0^+$.
Let $s=sigma+i omega$. The substitution becomes
$t to tau/s=tau/(sigma+i omega)=(tau(sigma-i omega))/(sigma^2+omega^2)$.
The integral is
$1/(sigma+i omega)^alpha (sigma-i omega)/(sigma^2+omega^2)int_(0^+)^(oo) tau^alpha e^(- tau) "d" tau=1/((sigma+iomega)^(1+alpha))int_(0^+)^(oo) tau^alpha e^(- tau) "d" tau=(Gamma(1+alpha))/(s^(1+alpha))$.
$t to tau/s=tau/(sigma+i omega)=(tau(sigma-i omega))/(sigma^2+omega^2)$.
The integral is
$1/(sigma+i omega)^alpha (sigma-i omega)/(sigma^2+omega^2)int_(0^+)^(oo) tau^alpha e^(- tau) "d" tau=1/((sigma+iomega)^(1+alpha))int_(0^+)^(oo) tau^alpha e^(- tau) "d" tau=(Gamma(1+alpha))/(s^(1+alpha))$.
"elgiovo":
Making the substitution $t to tau/s$, we obtain
$int_(0^+)^(oo)(tau/s)^(alpha)e^(-tau)("d"t)/s=1/(s^(1+alpha))int_(0^+)^(oo)tau^(alpha)e^(-tau)"d"tau$.
You're right, but that's true only for $s in RR^+$... instead Laplace transform is a function of complex variable.
Can you extend the result to all $s$ such as $Re(s) > 0$?
$ccL_u[t^(alpha)]=int_(0^+)^(oo)t^(alpha)e^(-st)"d"t$.
Making the substitution $t to tau/s$, we obtain
$int_(0^+)^(oo)(tau/s)^(alpha)e^(-tau)("d"t)/s=1/(s^(1+alpha))int_(0^+)^(oo)tau^(alpha)e^(-tau)"d"tau$.
This is a friendly integral, in fact for $Re[alpha+1]=alpha+1>0 to alpha> -1$ it is absolutely convergent and equals $Gamma(1+alpha)$. The final result is
$ccL_u[t^(alpha)]=(Gamma(1+alpha))/(s^(1+alpha))$.
BONUS QUESTION (I still don't know the answer): what if $alpha=-1$?
Making the substitution $t to tau/s$, we obtain
$int_(0^+)^(oo)(tau/s)^(alpha)e^(-tau)("d"t)/s=1/(s^(1+alpha))int_(0^+)^(oo)tau^(alpha)e^(-tau)"d"tau$.
This is a friendly integral, in fact for $Re[alpha+1]=alpha+1>0 to alpha> -1$ it is absolutely convergent and equals $Gamma(1+alpha)$. The final result is
$ccL_u[t^(alpha)]=(Gamma(1+alpha))/(s^(1+alpha))$.
BONUS QUESTION (I still don't know the answer): what if $alpha=-1$?
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