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I want to find the inverse Laplace transform of

$$F(s) = \frac{1}{\sqrt{s^2-a^2}}$$ preferably using the Bromwich integral: $$f(t) = \frac{1}{2\pi i}\int_{\beta -I \infty}^{\beta +i \infty}e^{st}F(s) ds $$

The problem is that the integrand has two branch points at $s=+a$ and $s=-a$ . (I've seen examples which have a branch point on the origin, that can be solved easily by excluding that point with an infinitesimal circle).

In order to be able to apply the Cauchy theorem to find the Bromwich integral on the original contour, the new contour should be like this, but I don't know how to perform the integration: enter image description here

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Consider the contour integral

$$\oint_C dz \frac{e^{t z}}{\sqrt{z^2-a^2}} $$

where $C$ is the contour drawn above, and $t \gt 0$. By Cauchy's theorem, this integral is zero. However, to evaluate the ILT, we need to evaluate all of the pieces of the contour integral. Thankfully, the OP has provided a diagram with such nice labels. Thus,

$$\int_{AB} dz \frac{e^{t z}}{\sqrt{z^2-a^2}} = \int_{\beta-i R}^{\beta+i R} ds \frac{e^{t s}}{\sqrt{s^2-a^2}}$$

$$\int_{BC} dz \frac{e^{t z}}{\sqrt{z^2-a^2}} = i R \int_{\pi/2}^{\pi} d\theta \, e^{i \theta} \frac{e^{t R e^{i \theta}}}{\sqrt{R^2 e^{i 2 \theta}-a^2}} $$

$$\int_{CD} dz \frac{e^{t z}}{\sqrt{z^2-a^2}} = \int_{-R}^{-a-i \epsilon} dx \frac{e^{t x}}{\sqrt{x^2-a^2}} $$

$$\int_{DE} dz \frac{e^{t z}}{\sqrt{z^2-a^2}} = i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{e^{t(-a+\epsilon e^{i \phi})}}{\sqrt{(-a+\epsilon e^{i \phi})^2-a^2}}$$

$$\int_{EF} dz \frac{e^{t z}}{\sqrt{z^2-a^2}} = \int_{-a+\epsilon}^{a-\epsilon} dx \frac{e^{t x}}{e^{i \pi/2} \sqrt{a^2-x^2}} $$

$$\int_{FG} dz \frac{e^{t z}}{\sqrt{z^2-a^2}} = i \epsilon \int_{\pi}^{-\pi} d\phi \, e^{i \phi} \frac{e^{t(a+\epsilon e^{i \phi})}}{\sqrt{(a+\epsilon e^{i \phi})^2-a^2}}$$

$$\int_{GH} dz \frac{e^{t z}}{\sqrt{z^2-a^2}} = \int_{a+\epsilon}^{-a-\epsilon} dx \frac{e^{t x}}{e^{-i \pi/2} \sqrt{a^2-x^2}} $$

$$\int_{HI} dz \frac{e^{t z}}{\sqrt{z^2-a^2}} = i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{e^{t(-a+\epsilon e^{i \phi})}}{\sqrt{(-a+\epsilon e^{i \phi})^2-a^2}}$$

$$\int_{IJ} dz \frac{e^{t z}}{\sqrt{z^2-a^2}} = \int_{-a-i \epsilon}^{-R} dx \frac{e^{t x}}{\sqrt{x^2-a^2}} $$

$$\int_{JA} dz \frac{e^{t z}}{\sqrt{z^2-a^2}} = i R \int_{\pi}^{3 \pi/2} d\theta \, e^{i \theta} \frac{e^{t R e^{i \theta}}}{\sqrt{R^2 e^{i 2 \theta}-a^2}} $$

OK, there's a lot there, but it's not nearly as bad as it looks. The integral over $AB$ will be $i 2 \pi$ times the ILT as $R \to \infty$. The integral over $BC$ vanishes in this limit because its magnitude is bounded by

$$\frac{R}{\sqrt{R^2-a^2}} \int_0^{\pi/2} d\theta \, e^{-t R \sin{\theta}} \le \frac{R}{\sqrt{R^2-a^2}} \int_0^{\pi/2} d\theta \, e^{-2 t R \theta/\pi} \le \frac{\pi}{2 t \sqrt{R^2-a^2}}$$

The integral over $JA$ vanishes for similar reasons. The integrals over $CD$ and $IJ$ cancel each other out. The integrals over $DE$, $HI$, and $FG$ vanish as $\epsilon \to 0$. Thus, in these limits, we may write the ILT as follows:

$$\int_{\beta-i \infty}^{\beta+i \infty} ds \frac{e^{t s}}{\sqrt{s^2-a^2}} - i 2 \int_{-a}^a dx \frac{e^{t x}}{\sqrt{a^2-x^2}} = 0$$

or

$$\frac1{i 2 \pi} \int_{\beta-i \infty}^{\beta+i \infty} ds \frac{e^{t s}}{\sqrt{s^2-a^2}} = \frac1{\pi} \int_{-a}^a dx \frac{e^{t x}}{\sqrt{a^2-x^2}} $$

We may evaluate the integral on the RHS as follows. Sub $x=a \cos{u}$; then the integral is equal to

$$\frac1{\pi} \int_0^{\pi} du \, e^{a t \cos{u}} = I_0(a t)$$

where $I_0$ is the modified Bessel function of the first kind of zeroth order. Thus,

$$\frac1{i 2 \pi} \int_{\beta-i \infty}^{\beta+i \infty} ds \frac{e^{t s}}{\sqrt{s^2-a^2}} = I_0(a t)$$

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  • $\begingroup$ Thanks a lot! I repeated the calculations for the case of imaginary $a$ too. But can the general case with complex $a$ be dealt with this way? The contour gets a little complicated with the branch points having nonzero imaginary and real parts. $\endgroup$ – user215721 Jan 9 '16 at 4:15
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    $\begingroup$ @user21572: In that case I would extend a branch to $a$ radially inward from the outer circular arc, along the line $\theta = \arg{a}$. $\endgroup$ – Ron Gordon Jan 9 '16 at 9:29
  • $\begingroup$ Sorry I had a problem with the argument for integration on $CD$ and $IJ$. Why these two cancel each other while $EF$ and $GH$ don't? $\endgroup$ – user215721 Jan 9 '16 at 15:14
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    $\begingroup$ @user215721: my apologies, I worked a little too quickly. The path above the branch cut has $\arg{(-1)} = \pi$, while that below has $\arg{(-1)} = -\pi$. Now, the arguments of the square roots on $CD$ and $IJ$ are positive, so that there is no need to use this fact and thus the integrals cancel. However, on $EF$ and $GH$, the arguments inside the sqrt are negative, so we do require the square root of $-1$, which differs according to branch. $\endgroup$ – Ron Gordon Jan 9 '16 at 15:50

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