Return to Answer

I have encountered the same problem! But let me emphasize, I am a geophysicist, not a mathematician!

My observations:

1. I do not find these (seemingly simple) transforms in most tables of Laplace transforms (not in Abramowitz and Stegun, for example). So, this is, to me, a "red flag".

2. I can do the complex-s-plane contour integral to obtain inverse LT of $\sqrt{s}$. It seems simple, there is a branch cut but no pole. The result is $\frac{-1}{2\sqrt{\pi}}t^{-3/2}$

3. Another way to evaluate the inverse LT of $\sqrt{s}$ is to use the convolution theorem and multiply by the LT of a Heaviside step function (1/s). The product is 1/$\sqrt{s}$ and the inverse LT of this IS found in standard tables, 1/$\sqrt{\pi t}$. If this is, then, differentiated with respect to t, we obtain $\frac{-1}{2\sqrt{\pi}}t^{-3/2}$

4. However, the forward Laplace transform of $t^{-3/2}$ does not converge! The problem, as might be suspected, is at t=0. This can be seen by calculating the indefinite integral and noting that the equation "blows up" at t = 0.

5. So this leaves us with what seems to me to be a paradox. The LT does not exist, but its inverse does! To me, this is also a red flag, but I don't know what it means.

I have encountered the same problem! But let me emphasize, I am a geophysicist, not a mathematician!

My observations:

1. I do not find these (seemingly simple) transforms in most tables of Laplace transforms (not in Abramowitz and Stegun, for example). So, this is, to me, a "red flag".

2. I can do the complex-s-plane contour integral to obtain inverse LT of $\sqrt{s}$. It seems simple, there is a branch cut but no pole. The result is $\frac{-1}{2\sqrt{\pi}}t^{-3/2}$

3. Another way to evaluate the inverse LT of $\sqrt{s}$ is to use the convolution theorem and multiply by the LT of a Heaviside step function (1/s). The product is 1/$\sqrt{s}$ and the inverse LT of this IS found in standard tables, 1/$\sqrt{\pi t}$. If this is, then, differentiated with respect to t, we obtain $\frac{-1}{2\sqrt{\pi}}t^{-3/2}$

4. However, the forward Laplace transform of $t^{-3/2}$ does not converge! The problem, as might be suspected, is at t=0. This can be seen by calculating the indefinite integral and noting that the equation "blows up" at t = 0.

5. So this leaves us with what seems to me to be a paradox. The LT does not exist, but its inverse does! To me, this is also a red flag, but I don't know what it means.

6. Subsequent comment: My thinking, now, is that the forward LT of $t^{-3/2}$ diverges because the LT is not exactly designed for simple transformation of quantities that are singular at t=0. One must properly accommodate the "pre-zero 0-" boundary condition with the "post-zero 0+" function. This might be addressed with "generalized functions" that can be discontinuous (But I don't, for the moment, know the details). Furthermore, I note that Abramowitz and Stegun do not layout mathematics that is explicitly compatible with this, see, for example section 29.1.1, where the lower limit of the LT is not very specifically identified as coming from the positive or negative direction. This might explain why they don't list the transforms of relevance to this discussion. Once the boundary condition at t=0 is handled, then the term that diverges in the LT of $t^{-3/2}$ might cancel, the remaining term is $\sqrt{s}$ -- thus there is the possibility of both forward LT compatibility with inverse LT. ;-)

I have encountered the same problem! But let me emphasize, I am a geophysicist, not a mathematician!

My observations:

1. I do not find these (seemingly simple) transforms in most tables of Laplace transforms (not in Abramowitz and Stegun, for example). So, this is, to me, a "red flag".

2. I can do the complex-s-plane contour integral to obtain inverse LT of $\sqrt{s}$. It seems simple, there is a branch cut but no pole. The result is $\frac{-1}{2\sqrt{\pi}}t^{-3/2}$

3. Another way to evaluate the inverse LT of $\sqrt{s}$ is to use the convolution theorem and multiply by the LT of a Heaviside step function (1/s). The product is 1/$\sqrt{s}$ and the inverse LT of this IS found in standard tables, 1/$\sqrt{\pi t}$. If this is, then, differentiated with respect to t, we obtain $\frac{-1}{2\sqrt{\pi}}t^{-3/2}$

4. However, the forward Laplace transform of $t^{-3/2}$ does not converge! The problem, as might be suspected, is at t=0. This can be seen by calculating the indefinite integral and noting that the equation "blows up" at t = 0.

5. So this leaves us with what seems to me to be a paradox. The LT does not exist, but its inverse does!

I have encountered the same problem! But let me emphasize, I am a geophysicist, not a mathematician!

My observations:

1. I do not find these (seemingly simple) transforms in most tables of Laplace transforms (not in Abramowitz and Stegun, for example). So, this is, to me, a "red flag".

2. I can do the complex-s-plane contour integral to obtain inverse LT of $\sqrt{s}$. It seems simple, there is a branch cut but no pole. The result is $\frac{-1}{2\sqrt{\pi}}t^{-3/2}$

3. Another way to evaluate the inverse LT of $\sqrt{s}$ is to use the convolution theorem and multiply by the LT of a Heaviside step function (1/s). The product is 1/$\sqrt{s}$ and the inverse LT of this IS found in standard tables, 1/$\sqrt{\pi t}$. If this is, then, differentiated with respect to t, we obtain $\frac{-1}{2\sqrt{\pi}}t^{-3/2}$

4. However, the forward Laplace transform of $t^{-3/2}$ does not converge! The problem, as might be suspected, is at t=0. This can be seen by calculating the indefinite integral and noting that the equation "blows up" at t = 0.

5. So this leaves us with what seems to me to be a paradox. The LT does not exist, but its inverse does! To me, this is also a red flag, but I don't know what it means.

I have encountered the same problem! But let me emphasize, I am a geophysicist, not a mathematician!

My observations:

1. I do not find these (seemingly simple) transforms in most tables of Laplace transforms (not in Abramowitz and Stegun, for example). So, this is, to me, a "red flag".

2. I can do the complex-s-plane contour integral to obtain inverse LT of sqrt(s). It seems simple, there is a branch cut but no pole. The result is -1/(2*sqrt(pi))*t^(-3/2)

3. Another way to evaluate the inverse LT of sqrt(s) is to multiply it by the LT of a Heaviside step function (1/s). The product is 1/sqrt(s) and the inverse LT of this IS found in standard tables, 1/sqrt(pit). If this is, then, differentiated with respect to t, we obtain -1/(2sqrt(pi))*t^(-3/2)

4. However, the forward Laplace transform of t^(-3/2) does not converge! The problem, as might be suspected, is at t=0. This can be seen by calculating the indefinite integral and noting that the equation "blows up" at t = 0.

5. So this leaves us with what seems to me to be a paradox. The LT does not exist, but its inverse does!

I have encountered the same problem! But let me emphasize, I am a geophysicist, not a mathematician!

My observations:

1. I do not find these (seemingly simple) transforms in most tables of Laplace transforms (not in Abramowitz and Stegun, for example). So, this is, to me, a "red flag".

2. I can do the complex-s-plane contour integral to obtain inverse LT of $\sqrt{s}$. It seems simple, there is a branch cut but no pole. The result is $\frac{-1}{2\sqrt{\pi}}t^{-3/2}$

3. Another way to evaluate the inverse LT of $\sqrt{s}$ is to use the convolution theorem and multiply by the LT of a Heaviside step function (1/s). The product is 1/$\sqrt{s}$ and the inverse LT of this IS found in standard tables, 1/$\sqrt{\pi t}$. If this is, then, differentiated with respect to t, we obtain $\frac{-1}{2\sqrt{\pi}}t^{-3/2}$

4. However, the forward Laplace transform of $t^{-3/2}$ does not converge! The problem, as might be suspected, is at t=0. This can be seen by calculating the indefinite integral and noting that the equation "blows up" at t = 0.

5. So this leaves us with what seems to me to be a paradox. The LT does not exist, but its inverse does!