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Where $s\in \mathbb{C}$. I assume that this would be pretty easily handled by solving it by definition, but I haven't taken courses in complex analysis yet. Also, I can't think of any nice property of the Laplace transform that would yield the answer. At second glance, a series represnation might do the trick, but then I remembered that s is a complex variable and thus the function can only be represented via (complex) Laurent series.

EDIT: I am looking for an approach different from what copper.hat suggested in the comments, namely using some of the nice properties of the transform.

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    $\begingroup$ You could try using $\ln(1+x^2) = \sum_{k=1}^\infty {x^{2n} \over n}$ for $|x| <1$. $\endgroup$ – copper.hat May 19 '15 at 16:40
  • $\begingroup$ @copper.hat Does that work with complex numbers? $\endgroup$ – Shemafied May 19 '15 at 16:58
  • $\begingroup$ Yes. ${}{}{}{}{}$ $\endgroup$ – copper.hat May 19 '15 at 16:59
  • $\begingroup$ Alrighty, then. $\endgroup$ – Shemafied May 19 '15 at 17:01
  • $\begingroup$ Maybe try $ln(1+ \frac{w^2}{s^2})=ln( \frac{1}{s^2} (s^2+w^2))=-ln(s^2)+ln(s^2+w^2)=-ln(s^2)+ln((s+iw)(s-iw))=-ln(s^2)+ln(s+iw)+ln(s-iw)$ and then use the linearity of the inverse laplace transform. $\endgroup$ – TheBluegrassMathematician May 21 '15 at 18:25
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Let's write down the expression for the inverse transform:

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, \log{\left (1+\frac{w^2}{s^2} \right )} e^{s t} $$

where $c$ is greater than the largest real part of all the singularities of the integrand, if any.

To compute the inverse transform above, consider the contour integral for $t \gt 0$:

$$\oint_{\gamma} dz \, \log{\left (1+\frac{w^2}{z^2} \right )} e^{z t} $$

where $\gamma$ is the contour illustrated below.

enter image description here

Note that we have deformed the contour so as to avoid the branch points at $z=\pm i w$. The outer circular arc has radius $R$ and the arcs about the branch points have radius $\epsilon$.

I will for the time being assure the reader that the integrals over the outer arcs vanish in the limit as $R \to \infty$, and the integrals over the inner arcs vanish in the limit as $\epsilon \to 0$. Therefore, we consider the contour integral as the limit as $R \to \infty$ and $\epsilon \to 0$.

To do this, we must look carefully at the behavior of the log term on the dog-bone portion of the contour. Note that

$$\log{\left (1+\frac{w^2}{z^2} \right )} = \log{\left (1+i \frac{w}{z} \right )}+\log{\left (1-i\frac{w}{z} \right )} $$

We take the log to be the principal branch, so that the branch cut occurs for the arg of the arguments of the log equal to $\pi$.

The reader should note that there is a fourfold split here: when $z$ is to the left or right of the imaginary axis, and when $z$ is above or below the real axis. In each of these cases, the log has a negative argument, but the log of the negative argument takes on a different value along each section of the dog-bone.

When $z$ is to the left and right of the imaginary axis we respectively parametrize $z=+i y$and $z=-i y$. However, we subtract $i 2 \pi$ from the log term upon crossing the imaginary axis and add $i 2 \pi$ upon crossing the real axis.

The contour integral is then equal to

$$\int_{c-i \infty}^{c+i \infty} ds \, \log{\left (1+\frac{w^2}{s^2} \right )} e^{s t} + i \int_0^w dy \, \left [\log{\left (\frac{w^2}{y^2}-1 \right )}+i \pi \right ] e^{i y t}\\ - i \int_w^0 dy \, \left [\log{\left (\frac{w^2}{y^2}-1 \right )}-i \pi \right ] e^{-i y t} - i \int_0^w dy \, \left [\log{\left (\frac{w^2}{y^2}-1 \right )}+i \pi \right ] e^{-i y t}\\+ i \int_w^0 dy \, \left [\log{\left (\frac{w^2}{y^2}-1 \right )}-i \pi \right ] e^{i y t}$$

Note that the log terms all cancel, and we are left with

$$\int_{c-i \infty}^{c+i \infty} ds \, \log{\left (1+\frac{w^2}{s^2} \right )} e^{s t} + i 2 \pi (i) (i 2) \int_0^w dy \, \sin{y t} $$

By Cauchy's theorem, the contour integral is zero. Therefore, we may immediately write down the inverse transform as

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, \log{\left (1+\frac{w^2}{s^2} \right )} e^{s t} = 2 \frac{1-\cos{w t}}{t} $$

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  • $\begingroup$ When you originally posted this answer I hadn't taken any complex analysis courses yet. However, now I have, and one thing that boggles me is how you came up with the different parametrizations for different sides of the branch cut. When there is only one cut, it is easy enough. Btw, your blog id awesome $\endgroup$ – Shemafied Dec 4 '15 at 8:19
  • $\begingroup$ W I have, and I am greatly confused on how you chose to parametrize the integral on different parts of the contours $\endgroup$ – Shemafied Dec 4 '15 at 8:36
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$$\ln\left( 1+\frac{w^2}{s^2}\right)=-\displaystyle\sum_{k=1}^\infty \frac{(-1)^k w^{2k}}{k s^{2k}} $$ The inverse Laplace transform of $\frac{1}{s^{2k}}$ is $\frac{t^{2k-1}}{(2k-1)!}$

So, the inverse laplace transform of $\displaystyle\sum_{k=1}^\infty \frac{(-1)^k w^{2k}}{k s^{2k}} $ is $-\displaystyle\sum_{k=1}^\infty \frac{(-1)^k w^{2k}t^{2k-1}}{k (2k-1)!} $ which is equal to :

$$-\frac{1}{t}\displaystyle\sum_{k=1}^\infty \frac{(-1)^k w^{2k}t^{2k}(2k)}{k (2k)(2k-1)!}=-\frac{2}{t}\displaystyle\sum_{k=1}^\infty \frac{(-1)^k w^{2k}t^{2k}}{ (2k)!}=-\frac{2}{t}\left(-1+\displaystyle\sum_{k=0}^\infty \frac{(-1)^k w^{2k}t^{2k}}{ (2k)!}\right)$$ Since $\displaystyle\sum_{k=0}^\infty \frac{(-1)^k (wt)^{2k}}{ (2k)!}=\cos(wt)$ ,

finally, the Laplace inverse is : $$\frac{2}{t}\left(1-\cos(wt)\right)$$

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  • $\begingroup$ I like both answers as of now, but I will wait until the last day of the bounty to reward someone. +1 for now $\endgroup$ – Shemafied May 22 '15 at 23:10
  • $\begingroup$ @ Shemafied : It's up to you to give the bounty to Ron Gordon. I am not following after rewards and I already got sufficient bounties. $\endgroup$ – JJacquelin May 24 '15 at 8:05

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