- #1
ralqs
- 97
- 1
According to the Wikipedia page, the inverse Laplace transform is
f(x) = \frac{1}{2 \pi i} \lim_{y\rightarrow \infty} \int_{x_0-iy}^{x_0+iy} F(s')e^{s'x}ds'
Something seems wrong though. If I were to take the Laplace transform this equation, I should get F(s) coming out of the right hand side. But when I try this, I get a stray factor of i:
\mathcal{L}(f(x))=\int_{-\infty}^{\infty}f(x)e^{-sx}dx = \frac{1}{2 \pi i} \lim_{y\rightarrow \infty} \int_{x_0-iy}^{x_0+iy} \int_{-\infty}^{\infty} F(s')e^{(s'-s)x}dxds' = \frac{1}{2 \pi i} \lim_{y\rightarrow \infty} \int_{x_0-iy}^{x_0+iy} F(s') [\int_{-\infty}^{\infty}e^{(s'-s)x}dx]ds'<br /> \frac{1}{2 \pi i} \lim_{y\rightarrow \infty} \int_{x_0-iy}^{x_0+iy} F(s') \cdot 2 \pi \delta (s'-s)ds'= -i F(s)
I would appreciate it if someone could identify my mistake. Thanks.
f(x) = \frac{1}{2 \pi i} \lim_{y\rightarrow \infty} \int_{x_0-iy}^{x_0+iy} F(s')e^{s'x}ds'
Something seems wrong though. If I were to take the Laplace transform this equation, I should get F(s) coming out of the right hand side. But when I try this, I get a stray factor of i:
\mathcal{L}(f(x))=\int_{-\infty}^{\infty}f(x)e^{-sx}dx = \frac{1}{2 \pi i} \lim_{y\rightarrow \infty} \int_{x_0-iy}^{x_0+iy} \int_{-\infty}^{\infty} F(s')e^{(s'-s)x}dxds' = \frac{1}{2 \pi i} \lim_{y\rightarrow \infty} \int_{x_0-iy}^{x_0+iy} F(s') [\int_{-\infty}^{\infty}e^{(s'-s)x}dx]ds'<br /> \frac{1}{2 \pi i} \lim_{y\rightarrow \infty} \int_{x_0-iy}^{x_0+iy} F(s') \cdot 2 \pi \delta (s'-s)ds'= -i F(s)
I would appreciate it if someone could identify my mistake. Thanks.
