Recent content by Talon44

So then, how do you evaluate the integral along the semicircular contour as R goes to infinity? Nvm, I see that an alternative shaped (rectangular) contour may be more effective for integrands containing ## e^{-x^2} ##. E.g., as shown here. I will try it and see.

@pasmith I apologize, I misunderstood what you were trying to accomplish with the simpler example – I did not understand why you were asking me to do other inverse transforms, but now it makes sense. In any case, thank you for taking the time to show that. I did not go through the exercise of...

Yes, it is straightforward to use the residue theorem with the Bromwich line integral to show that if $$ \bar{v}(p) = \frac{1}{\sqrt{p}+h} $$ Then $$ v(t) = \frac{2}{\pi} \int_0^{\infty} {\frac{e^{-tu^2} u^2 du}{h^2+u^2}} $$ Which has the analytical solution (according to Mathematica) $$...

Well, yes, it was already obvious that my result was not correct, which is why I am here. I know what the correct answer is - I'm trying to figure out how to derive it. It would be more useful if you could tell me what I'm doing wrong.

I appreciate your reply. At this point the transform is really just context. I am operating under the assumption that I have (inverse) transformed the solution to the subsidiary equation correctly. The integral form I am stuck at is already a transformed expression, one that is perfectly usable...

J Crank, The Mathematics of Diffusion 2nd Ed. Table 2.2. A similar transform (although with heat transfer variables) appears in Carslaw and Jaeger, Conduction of Heat in Solids. The transform I am practicing with is this: If $$ \bar{v}(p) = \frac{e^{-qx}}{q+h} $$ Then $$ v(t) =...

Just to be clear, the substitutions I originally made were: $$ \sin(ux) = \Im(e^{iux}) $$ $$ \cos(ux) = \Re(e^{iux}) $$ However, I tried it with the alternative substitutions you recommended and while the intermediate solutions to the residues for the sin and cos integrals were different, the...

Ok, I'll play along then. Here is what I did. For $$ \frac{2hD}{\pi} \int_0^\infty \frac{e^{-Du^2t}u \sin{ux}}{u^2+h^2} du $$ I substituted in ##e^{iux}## for the sine function, expecting only to use the imaginary part of the solution, so $$ f(z)= \frac{e^{-Dz^2t} e^{izx}z}{(z+ih)(z-ih)} $$...

Thank you for your response. Actually, what I'm really trying to do is show that $$ \frac{2hD}{\pi} \int_0^\infty \frac{e^{-Du^2t}u \sin{ux}}{u^2+h^2} du + \frac{2D}{\pi} \int_0^\infty \frac{e^{-Du^2t}u^2 \cos{ux}}{u^2+h^2} du $$ Has the analytical solution $$ \sqrt{\frac{D}{\pi t}}...

I am looking for a closed form solution to an integral of the form: $$ \int_0^\infty \frac{e^{-Du^2t}u \sin{ux}}{u^2+h^2} du $$ D, t, and h are positive and x is unrestricted. I have tried everything, integration by parts, substitution, even complex integration with residue analysis. I've...

I will need to spend some time wrapping my head around that, but I wanted to thank you for taking the time to reply. Solving heat/diffusion problems in cylindrical geometry requires manipulating these Bessel functions and I just don't have a lot of formal experience with them. I was doing pretty...

An infinite product representation of Bessel's function of the first kind is: $$J_\alpha(z) =\frac{(z/2)^\alpha}{\Gamma(\alpha+1)}\prod_{n=1}^\infty(1-\frac{z^2}{j_{n,\alpha}^2})$$ Here, the ##j_{n,\alpha}## are the various roots of the Bessel functions of the first kind. I found this...