I am facing the following problem. I know that the following relationship holds
$$A(\omega) = C + \int_{0}^{\infty} \frac{L(\tau)}{1 + i\omega \tau}\mathrm{d}\tau$$ where $C$ is a positive constant and $I$ is the imaginary unit. I would like to write the above as follows $$ \frac{1}{A(\omega) } = \frac{1}{C} + \int_{0}^{\infty} \frac{\hat{L}(\tau)}{1 + i\omega \tau}\mathrm{d}\tau$$ and I need to understand the relationship between $L$ and $\hat{L}$. The first attempt. Let me for simplicity denote $ l(\omega)=\int_{0}^{\infty} \frac{L(\tau)}{1 + i\omega \tau}\mathrm{d}\tau$, and $ \hat{l}(\omega)=\int_{0}^{\infty} \frac{\hat{L}(\tau)}{1 + i\omega \tau}\mathrm{d}\tau$: using the first equation to eliminate $A$ in the second, I got: $$\frac{1}{\hat{l}(\omega)} = -\frac{C^2}{l(\omega)} - C$$ but not made much progress afterwards in understanding how $L$ and $\hat{L}$ are related. As a matter of fact, I would struggle in an even simpler setting. Given $ l(\omega)=\int_{0}^{\infty} \frac{L(\tau)}{1 + i\omega \tau}\mathrm{d}\tau$, I could write $$\frac{1}{l(\omega)} = \frac{1}{\int_{0}^{\infty} \frac{L(\tau)}{1 + i\omega \tau}\mathrm{d}\tau}$$, but would not be able to go much further in expressing the r.h.s as an integral. The latter question made me think about how to re-write $$\frac{1}{a + b + c +...z} = \hat{a} + \hat{b} + \hat{c} + ... \hat{z}$$, where each "hatted" constant somehow depends on its counterpart, or all the other numbers $a,b,c..z$. Any hint would be the most appreciated, thanks
The question is asking if and how we can find $f(\tau)$, given that $$ F(\omega)=\int_0^\infty \frac{f(\tau)\,\mathrm{d}\tau}{1+i\omega\tau} . $$ Fix some constants $T>\varepsilon>0$, and consider $$ g_\varepsilon(T) = \int_{\Gamma_\varepsilon(T)} F(\omega)\,\mathrm{d}\omega , $$ where $\Gamma_\varepsilon(T)$ is the (positively oriented) circle of radius $\varepsilon$, centered at $\frac{i}T$. Proceeding formally, we get $$ g_\varepsilon(T) = \int_{\Gamma_\varepsilon(T)} \int_0^\infty \frac{f(\tau)}{1+i\omega\tau}\,\mathrm{d}\tau\,\mathrm{d}\omega = \int_0^\infty f(\tau) \left(\int_{\Gamma_\varepsilon(T)} \frac{\mathrm{d}\omega}{1+i\omega\tau}\right)\mathrm{d}\tau . $$ The inner integral is equal to $$ \int_{\Gamma_\varepsilon(T)} \frac{\mathrm{d}\omega}{1+i\omega\tau} = \begin{cases} \frac{2\pi}\tau&\left|\frac1\tau-\frac1T\right|<\varepsilon\\ 0&\left|\frac1\tau-\frac1T\right|>\varepsilon , \end{cases} $$ and hence $$ g_\varepsilon(T) = 2\pi\int_{\left|\frac1\tau-\frac1T\right|<\varepsilon} \frac{f(\tau)\,\mathrm{d}\tau}\tau . $$ For $\varepsilon>0$ small, we have $$ \int_{\left|\frac1\tau-\frac1T\right|<\varepsilon} \mathrm{d}\tau = 2\varepsilon T^2+O(\varepsilon^2) , $$ which leads to $$ g_\varepsilon(T) = 2\pi\cdot\frac{f(T)}T\cdot2\varepsilon T^2 + O(\varepsilon^2) = 4\pi\varepsilon T\,f(T) + O(\varepsilon^2) . $$ Therefore, we conclude $$ f(T) = \lim_{\varepsilon\to0}\frac{g_\varepsilon(T)}{4\pi\varepsilon T} = \lim_{\varepsilon\to0} \frac1{4\pi\varepsilon T} \int_{\Gamma_\varepsilon(T)} F(\omega)\,\mathrm{d}\omega . $$ As a final remark, we note that all the manipulations can be justified if we assume $f$ is continuous and has sufficient integrability at $0$ and $\infty$. Note also that $\Gamma_\varepsilon(T)$ does not have to be a circle; Any contour that wraps around the point $\frac{i}T$ that has a "radius" $\varepsilon$ should do.