Inverse Laplace transform of Gamma with branch cut
In solving a particular physical problem I have had to perform inverse Laplace transforms of sum and products of Gamma functions. Since my actual problem is complicated, I will state a simple example. Suppose:
$$F(s)=Gamma(s+a)-Gamma(s,b)\
Gamma(z)equivint_0^infty dt~ t^{z-1}e^{-t}\
Gamma(z,b)equivint_b^infty dt~ t^{z-1}e^{-t}$$
where $a,b>0$ are real.
The inverse Laplace transform is then:
$$f(t)=frac{1}{2pi i}int_Cds~e^{st}F(s)=frac{1}{2pi i}int_Cds~e^{st}(Gamma(s+a)-Gamma(s,b))$$
where $C$ is a vertical line in the complex $s$-plane such that all the singularities of $F(s)$ lie on the left of it.
$Gamma(s+a)$ has simple poles at $-(a+n),~nin{0,1,2,ldots}$ but no branch points. $Gamma(s,b)$ has a branch point at $0$ but has no poles. So I take my branch cut to be the positive real axis including origin. The rationale for this choice of the branch cut is that I want to evaluate the integral by summing the residues inside the closed contour shown in the figure above in the limit $Rtoinfty$. Following the logic in this post sum of residues of the integrand is simply: $sum_{n=0}^infty e^{-(a+n)t}(-1)^n/n!$. Therefore:
$$int_C ds~e^{st}F(s)+lim_{Rtoinfty}int_{tilde{C}}ds~e^{st}F(s)=sum_{n=0}^infty e^{-(a+n)t}frac{(-1)^n}{n!}$$
Assuming that everything I have done so far is correct then: How do I evaluate the second term on the LHS? Is it simply zero (how do I justify that)? Does the conclusion change if $F(s)=Gamma(s+a)Gamma(s,b)$?
P.S. I have read post1, post2, post3, but I still can't figure the answer to my question.
complex-integration branch-cuts inverselaplace
asked Nov 27 '18 at 16:58
Deep
371211
add a comment |
In solving a particular physical problem I have had to perform inverse Laplace transforms of sum and products of Gamma functions. Since my actual problem is complicated, I will state a simple example. Suppose:
$$F(s)=Gamma(s+a)-Gamma(s,b)\
Gamma(z)equivint_0^infty dt~ t^{z-1}e^{-t}\
Gamma(z,b)equivint_b^infty dt~ t^{z-1}e^{-t}$$
where $a,b>0$ are real.
The inverse Laplace transform is then:
$$f(t)=frac{1}{2pi i}int_Cds~e^{st}F(s)=frac{1}{2pi i}int_Cds~e^{st}(Gamma(s+a)-Gamma(s,b))$$
where $C$ is a vertical line in the complex $s$-plane such that all the singularities of $F(s)$ lie on the left of it.
$Gamma(s+a)$ has simple poles at $-(a+n),~nin{0,1,2,ldots}$ but no branch points. $Gamma(s,b)$ has a branch point at $0$ but has no poles. So I take my branch cut to be the positive real axis including origin. The rationale for this choice of the branch cut is that I want to evaluate the integral by summing the residues inside the closed contour shown in the figure above in the limit $Rtoinfty$. Following the logic in this post sum of residues of the integrand is simply: $sum_{n=0}^infty e^{-(a+n)t}(-1)^n/n!$. Therefore:
$$int_C ds~e^{st}F(s)+lim_{Rtoinfty}int_{tilde{C}}ds~e^{st}F(s)=sum_{n=0}^infty e^{-(a+n)t}frac{(-1)^n}{n!}$$
Assuming that everything I have done so far is correct then: How do I evaluate the second term on the LHS? Is it simply zero (how do I justify that)? Does the conclusion change if $F(s)=Gamma(s+a)Gamma(s,b)$?
P.S. I have read post1, post2, post3, but I still can't figure the answer to my question.
complex-integration branch-cuts inverselaplace
asked Nov 27 '18 at 16:58
Deep
371211
add a comment |
In solving a particular physical problem I have had to perform inverse Laplace transforms of sum and products of Gamma functions. Since my actual problem is complicated, I will state a simple example. Suppose:
$$F(s)=Gamma(s+a)-Gamma(s,b)\
Gamma(z)equivint_0^infty dt~ t^{z-1}e^{-t}\
Gamma(z,b)equivint_b^infty dt~ t^{z-1}e^{-t}$$
where $a,b>0$ are real.
The inverse Laplace transform is then:
$$f(t)=frac{1}{2pi i}int_Cds~e^{st}F(s)=frac{1}{2pi i}int_Cds~e^{st}(Gamma(s+a)-Gamma(s,b))$$
where $C$ is a vertical line in the complex $s$-plane such that all the singularities of $F(s)$ lie on the left of it.
$Gamma(s+a)$ has simple poles at $-(a+n),~nin{0,1,2,ldots}$ but no branch points. $Gamma(s,b)$ has a branch point at $0$ but has no poles. So I take my branch cut to be the positive real axis including origin. The rationale for this choice of the branch cut is that I want to evaluate the integral by summing the residues inside the closed contour shown in the figure above in the limit $Rtoinfty$. Following the logic in this post sum of residues of the integrand is simply: $sum_{n=0}^infty e^{-(a+n)t}(-1)^n/n!$. Therefore:
$$int_C ds~e^{st}F(s)+lim_{Rtoinfty}int_{tilde{C}}ds~e^{st}F(s)=sum_{n=0}^infty e^{-(a+n)t}frac{(-1)^n}{n!}$$
Assuming that everything I have done so far is correct then: How do I evaluate the second term on the LHS? Is it simply zero (how do I justify that)? Does the conclusion change if $F(s)=Gamma(s+a)Gamma(s,b)$?
P.S. I have read post1, post2, post3, but I still can't figure the answer to my question.
complex-integration branch-cuts inverselaplace
asked Nov 27 '18 at 16:58
Deep
371211
In solving a particular physical problem I have had to perform inverse Laplace transforms of sum and products of Gamma functions. Since my actual problem is complicated, I will state a simple example. Suppose:
$$F(s)=Gamma(s+a)-Gamma(s,b)\
Gamma(z)equivint_0^infty dt~ t^{z-1}e^{-t}\
Gamma(z,b)equivint_b^infty dt~ t^{z-1}e^{-t}$$
where $a,b>0$ are real.
The inverse Laplace transform is then:
$$f(t)=frac{1}{2pi i}int_Cds~e^{st}F(s)=frac{1}{2pi i}int_Cds~e^{st}(Gamma(s+a)-Gamma(s,b))$$
where $C$ is a vertical line in the complex $s$-plane such that all the singularities of $F(s)$ lie on the left of it.
$Gamma(s+a)$ has simple poles at $-(a+n),~nin{0,1,2,ldots}$ but no branch points. $Gamma(s,b)$ has a branch point at $0$ but has no poles. So I take my branch cut to be the positive real axis including origin. The rationale for this choice of the branch cut is that I want to evaluate the integral by summing the residues inside the closed contour shown in the figure above in the limit $Rtoinfty$. Following the logic in this post sum of residues of the integrand is simply: $sum_{n=0}^infty e^{-(a+n)t}(-1)^n/n!$. Therefore:
$$int_C ds~e^{st}F(s)+lim_{Rtoinfty}int_{tilde{C}}ds~e^{st}F(s)=sum_{n=0}^infty e^{-(a+n)t}frac{(-1)^n}{n!}$$
Assuming that everything I have done so far is correct then: How do I evaluate the second term on the LHS? Is it simply zero (how do I justify that)? Does the conclusion change if $F(s)=Gamma(s+a)Gamma(s,b)$?
P.S. I have read post1, post2, post3, but I still can't figure the answer to my question.
complex-integration branch-cuts inverselaplace
complex-integration branch-cuts inverselaplace
asked Nov 27 '18 at 16:58
Deep
371211
asked Nov 27 '18 at 16:58
Deep
371211
edited Nov 27 '18 at 18:08
asked Nov 27 '18 at 16:58
Deep
371211
asked Nov 27 '18 at 16:58
Deep
371211
asked Nov 27 '18 at 16:58
Deep
371211
371211
add a comment |
add a comment |
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