I am new to the distribution theory and have some difficulties to calculate curtain fourier transforms. Can you help me with $$\frac{e^{-xb}}{x+i0}$$
I got to the point
$$\lim_{\epsilon \to 0^{+}}\int\frac{e^{-xb+ixy}}{x+i\epsilon}$$
but cant get any further. I cannot find any integrable upper bound to justify the exchange of integral and limit nor calculate the integral itself. Any ideas?
Define one sided limit \begin{eqnarray} f(x+) & = & \lim_{t \rightarrow x^+} f(t) \ , \\ f(x-) & = & \lim_{t \rightarrow x^-} f(t) \end{eqnarray} for any point $x$ of $f$, where such a limit exists. These limits are applied for example in "Guenther & Lee: Partial Differential Equations of Mathematical Physics and Integral Equations" in the proof of Fourier inversion theorem. Define \begin{eqnarray} g(t) & = & \frac{1}{2}(\chi_{[0,\infty)}(t)+\chi_{(0,\infty)}(t)) \ , \\ f_\epsilon (t) & = & g(t)e^{-\epsilon t} \ . \end{eqnarray} Then \begin{equation} f(t) = \frac{1}{2}(f(t+)+f(t-)) \ . \end{equation} Define \begin{eqnarray} h_{\epsilon ,1}(t) = \frac{f_\epsilon (a-y-t+)-f_\epsilon (a-y+)}{-t} \ , t<0 \ , \\ h_{\epsilon ,2}(t) = \frac{f_\epsilon (a-y-t-)-f_\epsilon (a-y-)}{-t} \ , t>0 \ . \end{eqnarray} Then \begin{eqnarray} -th_{\epsilon ,1}(t-) & = & f_\epsilon (a-y-t+)-f_\epsilon (a-y+) \ , t \leq 0 \ , \\ -th_{\epsilon ,2}(t+) & = & f_\epsilon (a-y-t-)-f_\epsilon (a-y-) \ , t \geq 0 \ . \end{eqnarray} Note that $f_\epsilon(a-y-t+)$ is left differentiable at $a-y$ and $f_\epsilon(a-y-t-)$ right differentiable at $a-y$ even at $a-y=0$. Then $h(t-)$ and $h(t+)$ are continuous at $0$ and hence bounded near $0$. The difference quotients are bounded in the complements of the boundaries of $0$. Note that a boundary of $0$ exists within the definition of domain of $h_{\epsilon,k}$. Hence $h_{\epsilon ,k}$, $k \in \{1,2\}$ are bounded. Assuming $b = ia$, $a \in \mathbb{R}$ you might want to calculate \begin{eqnarray} \lim_{\epsilon \rightarrow 0^+} & PV & \int_{-\infty}^\infty \frac{e^{-bx+ixy}}{x+i\epsilon} dx = \lim_{\epsilon \rightarrow 0^+} -i PV \int_{-\infty}^\infty \frac{e^{-bx+ixy}}{-ix+\epsilon} dx \\ & = & \lim_{\epsilon \rightarrow 0^+} -i PV \int_{-\infty}^\infty \int_0^\infty e^{-\epsilon t} e^{ixt} dt e^{-i(a-y)x} dx = \lim_{\epsilon \rightarrow 0^+} -i PV \int_{-\infty}^\infty \int_0^\infty e^{-\epsilon t} e^{-ixt} dt e^{i(a-y)x} dx \\ & = & \lim_{\epsilon \rightarrow 0^+} -i \lim_{M \rightarrow \infty} \int_{-M}^M \int_{-\infty}^\infty f_\epsilon (t) e^{-ixt} dt e^{i(a-y)x} dx = \lim_{\epsilon \rightarrow 0^+} -i \lim_{M \rightarrow \infty} \int_{-\infty}^\infty \int_{-M}^M f_\epsilon (t) e^{i(a-y-t)x} dx dt \\ & = & \lim_{\epsilon \rightarrow 0^+} -i \lim_{M \rightarrow \infty} \int_{-\infty}^\infty \int_{-M}^M f_\epsilon (a-y-t) e^{itx} dx dt \\ & = & \lim_{\epsilon \rightarrow 0^+} -i \lim_{M \rightarrow \infty} \int_{-\infty}^0 f_\epsilon (a-y-t) \int_{-M}^M e^{itx} dx dt \\ & & + \lim_{\epsilon \rightarrow 0^+} -i \lim_{M \rightarrow \infty} \int_0^\infty f_\epsilon (a-y-t) \int_{-M}^M e^{itx} dx dt \\ & = & \lim_{\epsilon \rightarrow 0^+} -i \lim_{M \rightarrow \infty} \int_{-\infty}^0 (f_\epsilon (a-y-t+) - f_\epsilon (a-y+)) \int_{-M}^M e^{itx} dx dt + \lim_{\epsilon \rightarrow 0^+} -i \pi f_\epsilon (a-y+) \\ & & + \lim_{\epsilon \rightarrow 0^+} -i \lim_{M \rightarrow \infty} \int_0^\infty (f_\epsilon (a-y-t-) - f_\epsilon (a-y-)) \int_{-M}^M e^{itx} dx dt + \lim_{\epsilon \rightarrow 0^+} -i\pi f_\epsilon (a-y-) \\ & = & \lim_{\epsilon \rightarrow 0^+} -i \lim_{M \rightarrow \infty} \int_{-M_1}^0 -th_{\epsilon ,1}(t-) \int_{-M}^M e^{itx} dx dt + \lim_{\epsilon \rightarrow 0^+} -i \lim_{M \rightarrow \infty} \int_0^{M_2} -th_{\epsilon ,2}(t+) \int_{-M}^M e^{itx} dx dt \\ & & + \lim_{\epsilon \rightarrow 0^+} -2i \lim_{M \rightarrow \infty} \int_{-\infty}^{-M_1} (f_\epsilon (a-y-t+) - f_\epsilon (a-y+)) \frac{\sin(Mt)}{t} dt + \lim_{\epsilon \rightarrow 0^+} -i\pi f_\epsilon (a-y+) \\ & & + \lim_{\epsilon \rightarrow 0^+} -2i \lim_{M \rightarrow \infty} \int_{M_2}^\infty (f_\epsilon (a-y-t-) - f_\epsilon (a-y-)) \frac{\sin(Mt)}{t} dt + \lim_{\epsilon \rightarrow 0^+} -i\pi f_\epsilon (a-y-) \\ & = & \lim_{\epsilon \rightarrow 0^+} -2i \lim_{M \rightarrow \infty} \int_{-M_1}^0 -h_{\epsilon ,1}(t-) \sin(Mt) dt + \lim_{\epsilon \rightarrow 0^+} -2i \lim_{M \rightarrow \infty} \int_0^{M_2} -h_{\epsilon ,2}(t+) \sin(Mt) dt \\ & & + \lim_{\epsilon \rightarrow 0^+} -2i \lim_{M \rightarrow \infty} \int_{-\infty}^{-M_1} \frac{f_\epsilon (a-y-t+)}{t} \sin(Mt) dt \\ & & + \lim_{\epsilon \rightarrow 0^+} -2i \lim_{M \rightarrow \infty} -f_\epsilon (a-y+) \int_{-\infty}^{-M_1} \frac{\sin(Mt)}{Mt} Mdt \\ & & + \lim_{\epsilon \rightarrow 0^+} -2i \lim_{M \rightarrow \infty} \int_{M_2}^\infty \frac{f_\epsilon (a-y-t+)}{t} \sin(Mt) dt \\ & & + \lim_{\epsilon \rightarrow 0^+} -2i \lim_{M \rightarrow \infty} -f_\epsilon (a-y-) \int_{M_2}^\infty \frac{\sin(Mt)}{Mt} Mdt + \lim_{\epsilon \rightarrow 0^+} (-i\pi f_\epsilon (a-y+) -i \pi f_\epsilon (a-y-))\\ & = & \lim_{\epsilon \rightarrow 0^+} -2i \lim_{M \rightarrow \infty} -f_\epsilon (a-y+) \int_{-\infty}^{-M_1 M} \frac{\sin(t)}{t} dt \\ & & + \lim_{\epsilon \rightarrow 0^+} -2i \lim_{M \rightarrow \infty} -f_\epsilon (a-y-) \int_{M_2 M}^\infty \frac{\sin(t)}{t} dt + \lim_{\epsilon \rightarrow 0^+} (-i\pi f_\epsilon (a-y+) -i\pi f_\epsilon (a-y-)) \\ & = & -2\pi i \lim_{\epsilon \rightarrow 0^+} \frac{1}{2}(f_\epsilon (a-y+)+f_\epsilon (a-y-)) = -2\pi i \lim_{\epsilon \rightarrow 0^+} f_\epsilon(a-y) \\ & = & -2\pi i \lim_{\epsilon \rightarrow 0^+} g(a-y) e^{-\epsilon(a-y)} = -2\pi i g(a-y) = -2\pi i \frac{1}{2}(\chi_{[0,\infty)}(a-y)+\chi_{(0,\infty)}(a-y)) \ . \end{eqnarray} The first change of order of integration is ok by Fubini's theorem. The integrals from $-\infty$ to $0$ and from $0$ to $\infty$ of $\int_{-M}^M e^{itx} dx$ are equal to $\pi$. Riemann-Lebesgue -lemma is applied four times.
Ok, some narrative in English. The integral in the question is clearly $2\pi$ times Fourier transform at $-y$ because of the exponent $ixy$. We know that the Hilbert transform kernel is $\frac{1}{\pi t}$. We also know that its transfer function equals to $\frac{1}{2\pi i}\textrm{sgn}(\omega)$. I guess that the aim was to approximate the Hilbert kernel. However, note that the imaginary part $\frac{-\epsilon}{x^2+\epsilon^2}$ doesn't converge to $0$ as $\epsilon \rightarrow 0$. However, we keep the original question as assumption and transform $\frac{1}{x+i\epsilon}$. Now the problem is, that the function $\frac{1}{x+i\epsilon}$ isn't in $L^1$. Although we can write the function as a Fourier integral, we can't apply the Fourier inversion theorem that assumes that both the function and its transform are in $L^1$ (see W.Rudin: Functional Analysis). Hence we have to calculate the integral directly and apply suitable theorems.
The first thing is to ensure that the transform exists. For that purpose we use the Cauchy principal value and set $b=ia$, $a \in \mathbb{R}$. Otherwise the transform diverges at least at $0$. Then we write the fraction $\frac{1}{-ix+\epsilon}$ as a Fourier integral. The rest of the calculation follows the proof of Fourier series in "Stein & Shakarchi: Princeton lectures in analysis 2: Fourier analysis" and is applied to Fourier inversion theorem.
In the change of order of integration the integrand is absolutely integrable and Fubini's theorem can be applied. Then $f(a-y+)$ and $f(a-y-)$ are substracted and added under integrals. The integration of $\int_{-M}^M e^{itx} dx = \frac{2\sin(Mt)}{t}$, $t \neq 0$ returns to integration of $2\pi\textrm{sinc}(t)$, where $\textrm{sinc}(t) = \frac{\sin(\pi t)}{\pi t}$, $t \neq 0$ and $1$, $t = 0$, that is a known problem. This is also the way, that the value of the original integral is formed. The rest of the calculation is analysis of terms that converge to $0$. The analysis is done in two parts: from $0$ to $M_1$ and from $M_1$ to $\infty$. The integral of $f(a-y-t)-f(a-y-)$is equal to the integral of $f(a-y-t-)-f(a-y-)$ because they differ at most at one point $t = a-y$. The latter integral is analyzed also in two parts, this time separately for the two terms of the integrand. Then the Riemann-Lebesgue -lemma is applied to the part from $0$ to $M_1$ and the first term of the integral from $M_1$ to $\infty$. The last limit result used comes from the definition of limit, where the left hand side without absolute values $I_M-I$ is in the form $\int_{M_1 M}^\infty \frac{\sin(t)}{t} dt$. The rest of the calculation applies the definitions in the beginning of the text. Because $f_\epsilon$ has only one discontinuity at zero the verification of the equations goes without reference to semicontinuity properties of $\chi_{[0, \infty)}$ and $\chi_{(0,\infty)}$but with pointwise properties of the characteristic functions.
Assume $\varphi \in \mathcal{S}$. Define \begin{eqnarray} g_1(x) & = & |\varphi(x)| \ , x \in \mathbb{R} \ , \\ g_2(x) & = & \Bigg| \frac{\mathcal{F}\varphi(x)-\mathcal{F}\varphi(0)}{x} \Bigg| + \Bigg| \frac{\mathcal{F}\varphi(-x)-\mathcal{F}\varphi(0)}{-x} \Bigg| \ , x \in (0,M] \ , \ 0, x = 0 \ , \\ g_3(x) & = & \Bigg| \frac{\mathcal{F}\varphi(x)}{x} \Bigg| + \Bigg| \frac{\mathcal{F}\varphi(-x)}{-x} \Bigg| \ , x \in [M,\infty) \ . \end{eqnarray} Then $g_k$, $k \in \{1,2,3\}$ are clearly integrable. Define \begin{eqnarray} \Lambda_1(\varphi) & = & PV \int_{-\infty}^\infty \frac{1}{\pi x} \varphi(x) dx \ , \\ \Lambda_2(\varphi) & = & \int_{-\infty}^\infty \frac{1}{2\pi i} \textrm{sgn}(x) \varphi(x) dx \ . \end{eqnarray} You might want to calculate also \begin{eqnarray} \Lambda_2(\varphi) & = & \int_{-\infty}^\infty \frac{1}{2\pi i} \textrm{sgn}(x) \varphi(x) dx = \frac{1}{2\pi i} \int_{-\infty}^\infty \lim_{\epsilon \rightarrow 0^+} \textrm{sgn}(x) e^{-\epsilon |x|} \varphi(x) dx \\ & = & \frac{1}{2\pi i} \lim_{\epsilon \rightarrow 0^+} \int_{-\infty}^\infty \textrm{sgn}(x) e^{-\epsilon|x|} \varphi(x) dx \\ & = & \frac{1}{2\pi i} \lim_{\epsilon \rightarrow 0^+} \int_{-\infty}^\infty \textrm{sgn}(x) e^{-\epsilon|x|} \int_{-\infty}^\infty \mathcal{F}\varphi(t) e^{ixt} dt dx \\ & = & \frac{1}{2\pi i} \lim_{\epsilon \rightarrow 0^+} \int_{-\infty}^\infty \int_{-\infty}^\infty \textrm{sgn}(x) e^{-\epsilon|x|} e^{itx} \mathcal{F}\varphi(t) dx dt \\ & = & \frac{1}{2\pi i} \lim_{\epsilon \rightarrow 0^+} \int_{-\infty}^\infty \bigg( \int_{-\infty}^0 -e^{\epsilon x} e^{itx} dx + \int_0^\infty e^{-\epsilon x} e^{itx} dx \bigg) \mathcal{F}\varphi(t) dt \\ & = & \frac{1}{2\pi i} \lim_{\epsilon \rightarrow 0^+} \int_{-\infty}^\infty \bigg( -\frac{1}{it+\epsilon}-\frac{1}{it-\epsilon} \bigg) \mathcal{F}\varphi(t) dt \\ & = & \frac{1}{2\pi i} \lim_{\epsilon \rightarrow 0^+} \int_{-\infty}^\infty \bigg( \frac{-it + \epsilon - it - \epsilon}{-t^2-\epsilon^2} \bigg) \mathcal{F}\varphi(t) dt \\ & = & \frac{1}{\pi} \lim_{\epsilon \rightarrow 0^+} \int_{-\infty}^\infty \frac{t}{t^2+\epsilon^2} \mathcal{F}\varphi(t) dt \\ & = & \frac{1}{\pi} \lim_{\epsilon \rightarrow 0^+} \int_0^\infty \frac{t}{t^2+\epsilon^2} (\mathcal{F}\varphi(t) - \mathcal{F}\varphi(-t)) dt \\ & = & \frac{1}{\pi} \lim_{\epsilon \rightarrow 0^+} \int_0^M \frac{t}{t^2+\epsilon^2} ((\mathcal{F}\varphi(t)-\mathcal{F}\varphi(0))-(\mathcal{F}\varphi(-t)-\mathcal{F}\varphi(0))) dt \\ & & + \frac{1}{\pi} \lim_{\epsilon \rightarrow 0^+} \int_M^\infty \frac{t}{t^2+\epsilon^2} (\mathcal{F}\varphi(t)-\mathcal{F}\varphi(-t)) dt \\ & = & \frac{1}{\pi} \int_0^M \lim_{\epsilon \rightarrow 0^+} \frac{t}{t^2+\epsilon^2} ((\mathcal{F}\varphi(t)-\mathcal{F}\varphi(0))-(\mathcal{F}\varphi(-t)-\mathcal{F}\varphi(0))) dt \\ & & + \frac{1}{\pi} \int_M^\infty \lim_{\epsilon \rightarrow 0^+} \frac{t}{t^2+\epsilon^2} (\mathcal{F}\varphi(t)-\mathcal{F}\varphi(-t)) dt \\ & = & \frac{1}{\pi} \int_0^\infty \lim_{\epsilon \rightarrow 0^+} \frac{t}{t^2+\epsilon^2} (\mathcal{F}\varphi(t)-\mathcal{F}\varphi(-t)) dt \\ & = & \frac{1}{\pi} \lim_{\delta \rightarrow 0^+} \int_\delta^\infty \lim_{\epsilon \rightarrow 0^+} \frac{t}{t^2+\epsilon^2} (\mathcal{F}\varphi(t)-\mathcal{F}\varphi(-t)) dt \\ & = & \frac{1}{\pi} \lim_{\delta \rightarrow 0^+} \int_\delta^\infty \frac{1}{t} (\mathcal{F}\varphi(t)-\mathcal{F}\varphi(-t)) dt \\ & = & \frac{1}{\pi} \lim_{\delta \rightarrow 0^+} \bigg( \int_{-\infty}^{-\delta} \frac{1}{t} \mathcal{F}\varphi(t)) dt + \int_\delta^\infty \frac{1}{t} \mathcal{F}\varphi(t) dt \bigg) \\ & = & PV \int_{-\infty}^\infty \frac{1}{\pi t} \mathcal{F}\varphi(t) dt = \Lambda_1(\mathcal{F}\varphi) = \mathcal{F}\Lambda_1(\varphi) \ . \end{eqnarray} Hence \begin{eqnarray} \mathcal{F}\Lambda_1 = \Lambda_2 \ . \end{eqnarray} In the third equation the order of limit and integration is changed by Lebesgue dominated convergence theorem. In the fifth equation the order of integration is changed by Fubini's theorem. The other two changes of limit and integration are estalished by Lebesgue dominated convergence theorem.