# Tagged Questions

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### Fourier transform and inverse transorm

I have to prove that for $f\in L^1(\mathbb{R})$ $$\check{\hat{f}}=\hat{\check{f}},$$ where $$\hat{f}(\xi):=\int\limits_{\mathbb{R}}e^{-i\xi x}f(x)\mathrm{d}x$$ and ...
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### Inverting probability generating function via mellin transform substitution.

The pgf is defined as: $E(z^k)= \sum_{k=0}^{\infty} p(k)z^k$ which is a discretised version of the transform: $\widetilde{p}(z) =\int_{-\infty}^{\infty} z^k p(k) \, \mathrm{d}k$ The Mellin ...
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### Inverse of an integral transform

Suppose that in a certain domain of analyticity we're given a function $A(s)$ in terms of the integral : $$A(s)=\int_{0}^{\infty}\frac{a(t)}{t(t^{2}+s^{2})}dt$$ How can we recover $a(t)$?
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### How fast does the function $\displaystyle f(x)=\lim_{\epsilon\to0}\int_\epsilon^{\infty} \dfrac{e^{xt}}{t^t} \, dt$ grow?

Let $x$ be a positive real number and $f(x):=\lim_{\epsilon\to0}\int_\epsilon^{\infty} \dfrac{e^{xt}}{t^t} \, dt$. How fast does this function grow ? In other words can we find a good asymptote for ...
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### Steepest descent?

Here I would like to see the behavior of a function as an integral when its argument (which is a parameter in the integral) goes to zero. If I try to evaluate an integral ...
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### Problem with the Cauchy-Green transform

Let $u \in C_c(\mathbb C)$. Then it's Cauchy-Green transform $$\tilde u(z) = -\frac{1}{\pi} \int\limits_{\mathbb C} \frac{u(\xi)}{\xi-z} d\xi_R d\xi_I, \quad \xi = \xi_R + i \xi_I,$$ is ...
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### How to find the inverse mellin transform?

On the wikipedia page http://en.wikipedia.org/wiki/Mellin_transform the second formula is an integral transformation for the inverse Mellin transform. Being new to integral transforms, I wonder how ...
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### Analytic Continuation of a nowhere existing Mellin Transform

I'm trying to give sense to the (self-made) statement: The analytic continuation of $\int_0^\infty e^t\;t^{s-1}\;dt$ is holomorphic in $s=0$. At first sight this could seem completely ...
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### An inequality involving arctan of complex argument

I have the following conjecture: $$\text{Re}\left[(1+\text{i}y)\arctan\left(\frac{t}{1+\text{i}y}\right)\right] \ge \arctan(t), \qquad \forall y,t\ge0.$$ Which seems to be ...
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### Fourier series is to Fourier transform what Laurent series is to …?

Since the coefficients $$a_k = \frac1{2\pi i}\oint_C\frac{f(z)}{(z-c)^{k+1}}\,dz$$ for the Laurent series $$f(z)\Big|_{r\le|z|\le R} = \sum_{k=-\infty}^{\infty}a_k\cdot(z-c)^k$$ of a function ...
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### On using fourier transforms to solve the root of a convolution

In continuation of Lower bounds of laplace transform of characteristic functions. My question is: Can anyone point out where i'm going wrong in the derivation below. It's been a while ...
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### Condition for the inverse laplace transform of a function to exist and bromwich integral

Given any function, is there any way of determining from the nature of the function, if it is the laplace transform of a piecewise continuous function of exponential order? For e.g. say the function ...
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### Lambert transform of monomials

The Lambert transform of a function $f(x)$, is given by: $$\int_{0}^{\infty}\frac{f(x)}{e^{xt}-1}dx\;\;\;\;(\Re(t)>0)$$ We wish for a closed form of the transform : ...
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### arguing away - complex analysis

Probably a trivial question but I can't understand how to argue away the value of integrals in complex analysis. I am trying to find the inverse Laplace transform of $F(s)=\frac{1}{s(s+1)}$. The ...
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### Do integral transforms with meromorphic kernels always have analytic continuation?

Do integral transforms with meromorphic kernels always have analytic continuation ? I think so, but I do not know how to prove it. For clarity with analytic continuation I assume it was already ...
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### Proof of the Direct mapping Theorem for Mellin transform.

I cannot understand an integration in the proof of the Direct mapping Theorem for the Mellin transform. A statement of the Theorem, together with an outline of the standard proof, can be found at page ...
Let's consider a complex function that can be represented in the following form: $$K(z)=\int_{-\infty}^{\infty}A(\alpha)z^\alpha d\alpha$$ Writing $z=re^{i\theta}$, we get:  ...