What is the inverse of the integral transform in complex analysis?

[eljose]

Let be the integral transform Int(0,8=infinity)f(x)x^(-s-1)=g(s) what would be its inverse? i bet it would be the complex integral over the line Re(s)=c of
g(s)x^sds/s but i am not really sure..can somebody help me or give me a hint?..thanks.

 

[saltydog]

Let be the integral transform Int(0,8=infinity)f(x)x^(-s-1)=g(s) what would be its inverse? i bet it would be the complex integral over the line Re(s)=c of
g(s)x^sds/s but i am not really sure..can somebody help me or give me a hint?..thanks.

I don't understand your syntax. Is this your transform?

I\{f(x)\}=\int_0^\infty f(x)x^{-(s+1)}dx=g(s)

 

[matt grime]

Perhaps you'll heed my advice to learn some LaTeX now, eljose. I'm not being picky for the sake of being picky. There is even a thread in the physics forum about it and many on-line resources to learn from.

But, here, let f(x)=1

g(s) = \int_0^{\infty} x^{-(s+1)}dx

which isn't defined for many real s.

Actually, let f(x) be any polynomial in x and you don't get many real s for which it is defined.

 

[eljose]

Latex is very difficult forme to understand...if it were simpler i would learn it,in fact in Mathworld it appears the integral Int(1,8=Infinity)J(x)x^-s-1=LnR(s)/s where R(s) is the Riemman Zeta function,they solved it via the inverse transform
J(x)=Int(2-i8,2+i8)(x^s)LnR(s)/s you can check it at

http://mathworld.wolfram.com/RiemannPrimeCountingFunction.html

Yes saltydog this is just the integral transform i was using,know i would like to know if inverse exists

 

[shmoe]

This is just the Mellin transform with -s instead of the usual s. You should be able to find references on this just about anywhere, it's pretty standard stuff.

I've suggested it before too, learn some LaTeX already. It's really not difficult to learn basics if you've got some examples to work with, and on this board there are plenty. If you click on the integrals in this post it gives the code used to generate them, it's then rather simple to modify them.

 

[saltydog]

Seems to be this:

M\{f(x)\}=\int_1^\infty f(x)x^{-(s+1)}dx=F(s)

And the inverse transform is:

M^{-1}\{F(s)\}=\frac{1}{2\pi i}\int_{c+i\infty}^{c-i\infty} x^sF(s)ds=f(x)

By the residue theorem:

f(x)=\sum_{i=1}^k Res\{x^{s_i}F(s_i)\}

Where s_i is a pole and k is the total number of poles. Can someone verify this please?

 

[saltydog]

Seems to be this:

M\{f(x)\}=\int_1^\infty f(x)x^{-(s+1)}dx=F(s)

And the inverse transform is:

M^{-1}\{F(s)\}=\frac{1}{2\pi i}\int_{c+i\infty}^{c-i\infty} x^sF(s)ds=f(x)

By the residue theorem:

f(x)=\sum_{i=1}^k Res\{x^{s_i}F(s_i)\}

Where s_i is a pole and k is the total number of poles. Can someone verify this please?

Alright, I used the following transform pair:

f(x)=\frac{1}{x^2}

F(s)=\frac{1}{2+s}

Seems to work according to the transform pair if I'm doing the complex contour correctly. Can anyone check. It's ok if you guys are busy with work and school. I have time and just having fun. You know, there's some very interesting questions presented here.

 

[shmoe]

The integral in the Mellin transform runs from 0 to infinity, so you'll run into convergence issues with your choice of f(x). See http://mathworld.wolfram.com/MellinTransform.html

Note that the example involving the Zeta function has this J(x) function, which is zero if x<2, so there's no harm in changing the lower limit to 1.

Be careful in how you evaluate the complex integral using residues. Which residues are you including? When you apply the residue theorem to a bounded countour (say the rectangle c+iT,d+iT,d-iT, and c-iT) and let that contour wander off to infinity (say T->infinity and d->+ or - infinity), you have to be sure that the parts of the contour you aren't interested in don't contribute (or if they do, you need to evaluate or at least bound this contribution somehow). Keep in mind that sometimes the integral involved in the inverse transform is only conditionally convergent, and by the infinite endpoints we mean the Cauchy Principle value.

 

[saltydog]

The integral in the Mellin transform runs from 0 to infinity, so you'll run into convergence issues with your choice of f(x). See http://mathworld.wolfram.com/MellinTransform.html

Note that the example involving the Zeta function has this J(x) function, which is zero if x<2, so there's no harm in changing the lower limit to 1.

Be careful in how you evaluate the complex integral using residues. Which residues are you including? When you apply the residue theorem to a bounded countour (say the rectangle c+iT,d+iT,d-iT, and c-iT) and let that contour wander off to infinity (say T->infinity and d->+ or - infinity), you have to be sure that the parts of the contour you aren't interested in don't contribute (or if they do, you need to evaluate or at least bound this contribution somehow). Keep in mind that sometimes the integral involved in the inverse transform is only conditionally convergent, and by the infinite endpoints we mean the Cauchy Principle value.

Thanks. I'll regroup.
Salty

 

[saltydog]

Thanks. I'll regroup.
Salty

Jesus. Even relatively simple Mellin transforms turn out to be tough to evaluate . . . the more I know, the less I know I know. . .
Might spend some time with it though.

Salty