Complex variable line integral...

**chisigma** · September 4th 2009, 12:07 AM

My problem is computing the function of real variable $y$ ...

$\delta (y) = \frac{1}{2\pi i} \int_{c-i \infty}^{c+i \infty} \frac{y^{s}}{s} \cdot ds$ (1)

... where $c >0$ is real, i.e. the integral of the function $\frac{y^{s}}{2\pi i s}$ along the line defined as $Re(s)=c >0$ . In particular is important for me to undestand why the integral (1) doesn't depend from $c$ ...

Any help will be greatly appreciated!...

Kind regards

$\chi$ $\sigma$

**Opalg** · September 4th 2009, 04:35 AM

Originally Posted by chisigma

My problem is computing the function of real variable $y$ ...

$\delta (y) = \frac{1}{2\pi i} \int_{c-i \infty}^{c+i \infty} \frac{y^{s}}{s} \cdot ds$ (1)

... where $c >0$ is real, i.e. the integral of the function $\frac{y^{s}}{2\pi i s}$ along the line defined as $Re(s)=c >0$ . In particular is important for me to understand why the integral (1) doesn't depend from $c$ ...

Let $c_1>c_2>0$ and consider the integral $\int_{\Gamma}\frac{y^s}s\,ds$ , where $\Gamma$ is the rectangular contour with vertices at $c_1\pm iR$ and $c_2\pm iR$ (for some large real number R). The only singularity of the integrand is at 0 (and hence not inside the contour) so the integral is zero. The components of the integral along the top and bottom segments of the contour will be small for large R, so the integrals along the vertical segments will be similar, and in fact will coincide in the limit as R→∞. Hence $\int_{c_1-i\infty}^{c_1+i\infty}\frac{y^s}s\,ds = \int_{c_2-i\infty}^{c_2+i\infty}\frac{y^s}s\,ds$ .

**chisigma** · September 4th 2009, 05:14 AM

Thanks to Opalg and his allways precise answers!

...

... by the time I found, after some uncertainties, the way to compute the integral...

$\delta(y)= \frac{1}{2\pi i} \int_{c - i \infty} ^ {c + i \infty} \frac {y^{s}}{s} \cdot ds$ (1)

The decisive step has been to remember the nice 'Complex inversion formula of the Laplace Transform' found by the British mathematician Thomas John I'Anson Bromwich about a century ago...

$\varphi(t) = L^{-1} \{f(s)\}= \frac{1} {2\pi i} \int_{c-i \infty}^{c+i \infty} e^{st}\cdot f(s) \cdot ds$ (2)

Setting $t=\ln y$ and $f(s)= \frac{1}{s}$ in (2) we obtain the (1) so that is...

$\delta (y) = u(\ln y)$ (3)

... where $u(*)$ is the so called 'Heaviside step function'. So the poblem seems to be solved and is...

$\delta(y)=\left\{\begin{array}{cc}0,&\mbox{ if }<br /> 0<y<1\\1, & \mbox{ if } y>1\end{array}\right.$ (4)

... all right!... wonderfull!

... but

... what about $\delta(1)$ ?

...

Kind regards

$\chi$ $\sigma$

P.S. According to...

http://en.wikipedia.org/wiki/Heaviside_step_function

... the question of the value of $u(0)$ is a little controversial... a good compromise seems to be $u(0)=\frac{1}{2}$ , so that is $\delta(1) = \frac{1}{2}$ ... mah!...

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