# Tagged Questions

This covers all transformations of functions by integrals, including but not limited to the Radon, X-Ray, Hilbert, Mellin transforms. Use (wavelet-transform), (laplace-transform) for those respective transforms, and use (fourier-analysis) for questions about the Fourier and Fourier-sine/cosine ...

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### Can inversion integral of characteristic functions on a finte interval be bounded?

For a real-valued uni-variate r.v. $X$, with pdf $f(x)$ and absolute integrable cf $\varphi(t)$, we have the following transform:$$2\pi f(x)=\int_{-\infty}^{\infty}e^{-itx}\varphi(t)\,dt.$$ However, I ...
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### Transformation of spherical means

I just have a small question: It is always stated that $$\frac{1}{n \alpha (n)r^{n-1}}\int_{\partial B(0,r)} u(x) \text{d}x = \frac{1}{n \alpha (n)}\int_{\partial B(0,1)} u(rx) \text{d}x$$ Now I ...
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### Fourier transform spherically symmetric function with complex constant

In Gradshteyn's section 17.24 on Fourier transform pairs for spherically symmetric functions, the third entry relates $\frac{e^{-ar}}{r}$ and $\sqrt{\frac{2}{\pi}}\frac{1}{(a^2 + k^2)^2}$. I think ...
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### Computing integral by using variable transformation

Let $I := \int_{(0,1)^2}\frac{1}{1-xy}\, d\lambda^2 (x,y)$. Can someone help me to determine $I$ only buy using the transformations $u=\frac{1}{2} (y+x)$ and $v=\frac{1}{2} (y-x)$? I don't know how ...
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### Fourier inverse/convolution problem

I'm struggling to do part (b) of this problem. I do not know how to start: I'm trying to use the inversion formula, but I don't know what to do with the $e^{|s|}$ part (the other one is the laplace ...
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### Riesz transform does not preserve continuity

I've read somewhere that the Riesz operator $R_j$ defined by $$R_j f(t) := c(n) \, \text{pv} \int_{\mathbb{R}^n} \frac{x_j}{|x|^{n+1}} f(t-x) \, dx$$doesn't preserve the continuity, but I can't ...
I am trying to find the function whose laplace transform is below using the complex inversion formula: $$f(s)= \frac{se^s}{(s-2)^3}$$ My attempt below seems to be giving me the wrong answer but I'm ...