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| bio | website | |
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| age | ||
| visits | member for | 1 year, 4 months |
| seen | May 16 at 2:52 | |
| stats | profile views | 260 |
I am a physics undergrad with interests in theoretical physics and mathematics.
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May 13 |
revised |
Show that the projection map is Orientation preserving iff n is even spelling mistake in title |
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May 13 |
suggested | suggested edit on Show that the projection map is Orientation preserving iff n is even |
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May 13 |
revised |
Show that the projection map is Orientation preserving iff n is even edited title |
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May 13 |
suggested | suggested edit on Show that the projection map is Orientation preserving iff n is even |
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May 8 |
comment |
Graduate research project in stochastic programming . I think this question is not suited for this site(?). You may however post it on Academica SE: academia.stackexchange.com |
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May 6 |
comment |
Prove there is no element of order 6 in a simple group of order 168 Why does $n_2=21$, mean there is no element of order 6? |
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May 6 |
awarded | Caucus |
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May 6 |
accepted | Laurent expansion of $\csc^2(\frac{\pi}{z})$ about $\frac{1}{3}$ for $|z-\frac{1}{3}| \lt \frac{1}{12}$ |
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Apr 30 |
accepted | Problem evaluating an improper integral $\int_0^{\infty} \frac{(\sin{2x}-2x\cos{2x})^2}{x^6}$ using fourier transform |
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Apr 30 |
comment |
Problem evaluating an improper integral $\int_0^{\infty} \frac{(\sin{2x}-2x\cos{2x})^2}{x^6}$ using fourier transform Thanks! I missed a term in the second integral. I am really clumsy in my calculations. Sorry for the trouble. |
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Apr 30 |
comment |
Problem evaluating an improper integral $\int_0^{\infty} \frac{(\sin{2x}-2x\cos{2x})^2}{x^6}$ using fourier transform At first glance, I am flabbergasted. The first term $\int_o^{\infty}a^2 \cos{\xi x} dx=a^2\frac{\sin{a \xi}}{\xi}=\frac{a^2 \xi ^2 \sin{a \xi} }{\xi ^3}$, which is clearly missing from your answer, and clearly present in mine. The second part done by parts, should't cancel it out. Thanks a lot for the trouble though. :) |
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Apr 30 |
comment |
Problem evaluating an improper integral $\int_0^{\infty} \frac{(\sin{2x}-2x\cos{2x})^2}{x^6}$ using fourier transform Hi. Should I have taken the sine fourier transform? My sine and cos are interchanged. But the function is even, so the sine part would be zero right? |
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Apr 30 |
comment |
Problem evaluating an improper integral $\int_0^{\infty} \frac{(\sin{2x}-2x\cos{2x})^2}{x^6}$ using fourier transform @BarackObama: I didn't know the plancherel's formula, so I looked it up on the internet, which correct me in I am wrong is $\int_{\mathbb{R}}||f(x)||^2 \, dx = \int_{\mathbb{R}}||\hat{f(\xi)}||^2 \, d {\xi}$. But in my problem, the integrand is not the square of the entire fourier transform. There is an extra term, which I won't be able to integrate after squaring. How do I do it? |
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Apr 30 |
revised |
Problem evaluating an improper integral $\int_0^{\infty} \frac{(\sin{2x}-2x\cos{2x})^2}{x^6}$ using fourier transform added 32 characters in body; edited title |
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Apr 30 |
asked | Problem evaluating an improper integral $\int_0^{\infty} \frac{(\sin{2x}-2x\cos{2x})^2}{x^6}$ using fourier transform |
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Apr 30 |
accepted | Evaluating improper integrals using laplace transform |
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Apr 30 |
comment |
Evaluating improper integrals using laplace transform Awesome! Thanks. |
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Apr 30 |
asked | Evaluating improper integrals using laplace transform |
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Apr 30 |
comment |
Condition for the inverse laplace transform of a function to exist and bromwich integral Assuming your $t$ is my $x$, I don't see why $\frac{2e^{iR}}{x^3}$ would be zero, if $x$ is non zero. Have I dont the contour integral right? Here there are no singularities so my $a$ can be anything, and my contour is the straight line parallel to the imaginary axis. Also, do you have a answer to my first query. Is there any theorem, which places some restriction on my $F(s)$ so that it has a inverse laplace transform. |
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Apr 30 |
asked | Condition for the inverse laplace transform of a function to exist and bromwich integral |
