Let $0 < x < 1$, I have to compute this Laplace transform:

$$ f(x) = \int_0^\infty dt \; e^{-t/g} \; \frac{1}{\sqrt{1 - 2 t x}} $$

I am not 100% this interal is defined. If $t > \frac{1}{x}$ this integral is imaginary. If we ignore that issue, let $s = tx$ and $ds = x \, dt$.

$$ \frac{1}{x} \int_0^\infty ds \; e^{-s/gx} \; \frac{1}{\sqrt{1 - 2 s}} $$

This is not going to work since $1-2s < 0$ for $s > \frac{1}{2}$ and the square root becomes negative. In fact as $s$ goes from $0$ to infinity, $\frac{1}{\sqrt{1-2s}}$ goes from $1 \to \infty$ along the real axis and $\infty \to 0$ along the imaginary axis.

The circle is finished by another function: $\frac{1}{\sqrt{1+2s}}$ which moves from $1 \to 0$ along the real axis.

How about if we tried a complex integral:

$$\int_0^\infty dt \; e^{-t/g} \left[ \frac{1}{\sqrt{1 - (2t+i\epsilon)x}} - \frac{-i e^{\frac{1}{2g}} }{\sqrt{1 - (-2t + i\epsilon)x}} \right]$$

Can this be expressed as a complex integral with a simple pole? Then I can take the residue. The $\epsilon > 0$ is just to avoid branch cut issues.

The answer should simply be $e^{-\frac{x}{2g}}$

  • $\begingroup$ i think you have to pull ur path of integration slightly below or above the cut for $t>x/2$, therefore shifting $t\rightarrow t\pm i \delta$, with $\delta\rightarrow 0_+$ to make the integral well defined. Which way u choose may depend on the actual context where this integral shows up $\endgroup$ – tired Jan 25 '16 at 22:52
  • $\begingroup$ i also suspect that the result will somehow contain an error function. where does this result come from? $\endgroup$ – tired Jan 25 '16 at 23:02
  • $\begingroup$ @tired you get the error function if you don't add any correcting terms wolframalpha.com/input/… that's why I started making stuff up. it comes from a physics paper $\endgroup$ – cactus314 Jan 25 '16 at 23:03
  • $\begingroup$ the errorfunction stems from the part of the integral $0<t< x/2$. I think what they are doing is that they travel a path from $[0, \infty + i \delta]$ encirceling the branch cut at infinity and come back along the way $[ \infty - i \delta,0]$. In this case the part which produces the errofunction cancels out (because the function is single valued there), and we are left with a well defined integral over the branch cut which should yield the correct result (but im wondering why there is no $1/2gx$) in the exponent. $\endgroup$ – tired Jan 25 '16 at 23:12
  • $\begingroup$ is there any chance this integral occurs in the context of intstation calculations in QFT? I saw something like this before $\endgroup$ – tired Jan 25 '16 at 23:14

Ok, as it stands the integral is not well defined. To give it a meaning we somehow need to regularize it. One way of doing that is to define the value of the integral as ($\delta\rightarrow 0_+$)

$$ I\equiv\frac{1}{2}\left(\int_0^{\infty+i\delta}e^{-t/g}\frac{1}{\sqrt{1-2 t x}}dt+\int_{\infty-i\delta}^0e^{-t/g}\frac{1}{\sqrt{1-2 t x}}dt\right) $$

This somehow resembles the definition of the principal value commonly used in complex analysis $P\frac{1}{x}=\frac{1}{2}\left(\frac{1}{x+i\delta}+\frac{1}{x-i\delta}\right)$ extended to the case of a non isolated singularity (a branch cut). Furthermore we may interpret this as employing the standard time ordering procedure of Quantum Field Theory (The above is somehow equivalent to add a retared and advanced ''propagator'' which gives a ''time orderd'' propagator) Now, because the square root is well defined for $0<t<x/2$ this part of the integrals will cancel out for sure, and we are left with $$ I=\frac{1}{2}\left(\int_{x/2+i\delta}^{\infty+i\delta}e^{-t/g}\frac{1}{\sqrt{1-2 t x}}dt-\int_{x/2-i\delta}^{\infty-i\delta}e^{-t/g}\frac{1}{\sqrt{1-2 t x}}dt\right) $$

Taking the principal branch of the logarithm we may obtain

$$ I=i\int_{x/2}^{\infty}e^{-t/g}\frac{1}{\sqrt{2xt-1}} $$

This integral is quiet standard to solve taking $2xt-1=q^2$ we obtain

$$ I=\frac{i}{x}e^{-1/(2gx)}\int_{0}^{\infty}e^{-q^2/(2gx)}dq $$

which is a standard Gaussian integral yielding $$ I=\frac{i \sqrt{\pi g}}{\sqrt{2x}}e^{-1/(2gx)} $$

  • $\begingroup$ Yeah this could be the one $\endgroup$ – cactus314 Jan 26 '16 at 3:39

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