- #1
SAT999
- 1
- 0
hello. I have transformed the Laplace transform into the infinite series by repeatedly using integration by parts.
What is this infinite series? may be Laplace transform series, or only an infinite series without name?
<br /> L(t)= \int_{t}^{\infty}\frac{f(t)}{e^{st}} dt =-0 + <br /> <br /> \frac{1}{e^{st}}\sum_{n=0}^{\infty}\frac{f^{(n)}(t)}{s^{(n+1)}} <br />
<br /> <br /> IL(L(t))= \frac{1}{2\Pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}e^{st}L(t) ds=\frac{1}{2\Pi <br /> <br /> i}\oint_{c}e^{st}L(t) ds = \frac{1}{2\Pi i}\sum_{n=0}^{\infty}\oint_{c}\frac{f^{(n)}(t)}{s^{(n+1)}} <br /> <br /> ds<br />
<br /> = \frac{1}{2\Pi i}f(t)\oint_{c}\frac{1}{s} ds = \frac{1}{2\Pi i}f(t)\int_{0}^{2\Pi}\frac{is}{s} <br /> <br /> d\Theta = f(t)<br />
What is this infinite series? may be Laplace transform series, or only an infinite series without name?
<br /> L(t)= \int_{t}^{\infty}\frac{f(t)}{e^{st}} dt =-0 + <br /> <br /> \frac{1}{e^{st}}\sum_{n=0}^{\infty}\frac{f^{(n)}(t)}{s^{(n+1)}} <br />
<br /> <br /> IL(L(t))= \frac{1}{2\Pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}e^{st}L(t) ds=\frac{1}{2\Pi <br /> <br /> i}\oint_{c}e^{st}L(t) ds = \frac{1}{2\Pi i}\sum_{n=0}^{\infty}\oint_{c}\frac{f^{(n)}(t)}{s^{(n+1)}} <br /> <br /> ds<br />
<br /> = \frac{1}{2\Pi i}f(t)\oint_{c}\frac{1}{s} ds = \frac{1}{2\Pi i}f(t)\int_{0}^{2\Pi}\frac{is}{s} <br /> <br /> d\Theta = f(t)<br />
