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I wannnaaaaa publishhhhh aaaa bbbboookkkkkkkkkk :-)))))))


Jan
23
awarded  Good Question
Jan
22
awarded  Nice Question
Jan
13
awarded  Popular Question
Jan
7
comment Inequality involving sides of a triangle
See the deleted comment (pls do not share it, I wanna add it to my book) :-)
Jan
7
comment How to evaluate the integral $\int_1^{\infty}[u^\alpha-(u-1)^{\alpha}]^2du$?
Nicely done (+1)
Jan
3
awarded  Nice Question
Jan
1
comment $f \circ g =\operatorname{ id}$ and $g \circ f \neq \operatorname{id}$?
See above the result.
Dec
29
comment Explanation needed on simple definite integral homework
@robjohn see the deleted message
Dec
28
revised Integration $\int_{m}^\infty {x}{\sqrt{x^2 - m^2}}e^{-\beta x} dx$
added 137 characters in body
Dec
28
comment Integration $\int_{m}^\infty {x}{\sqrt{x^2 - m^2}}e^{-\beta x} dx$
@MycrofD Use the recurrence relation $$K_v(z)=K_{v-2}(z)+2\frac{v-1}{z}K_{v-1}(z)$$
Dec
28
comment Integration $\int_{m}^\infty {x}{\sqrt{x^2 - m^2}}e^{-\beta x} dx$
@MycrofD I used Mathematica to differentiate with respect to $\beta$. The answer is checked numerically and it's correct. When I'll have some time I'll expand the answer.
Dec
28
comment Integration $\int_{m}^\infty {x}{\sqrt{x^2 - m^2}}e^{-\beta x} dx$
@MycrofD Before posting this answer, I discussed with Hippalectryon in a chat the formula he posted it above. I used it and differentiated it properly with respect to $\beta$ such that I got the integral you wanted to evaluate. I don't think you can simplify further the last expression.
Dec
27
revised Integration $\int_{m}^\infty {x}{\sqrt{x^2 - m^2}}e^{-\beta x} dx$
added 27 characters in body
Dec
27
comment Integration $\int_{m}^\infty {x}{\sqrt{x^2 - m^2}}e^{-\beta x} dx$
@alexjo Yeah, it's just a different way of writing your result that looks simpler.
Dec
27
revised Integration $\int_{m}^\infty {x}{\sqrt{x^2 - m^2}}e^{-\beta x} dx$
added 136 characters in body
Dec
27
answered Integration $\int_{m}^\infty {x}{\sqrt{x^2 - m^2}}e^{-\beta x} dx$
Dec
22
awarded  Necromancer
Dec
20
answered Closed form of $\int_0^1\int_0^1\int_0^1\frac{\left(1-x^y\right)\left(1-x^z\right)\ln x}{(1-x)^3}\,\mathrm dx\;\mathrm dy\;\mathrm dz$
Dec
20
comment Putnam definite integral evaluation $\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx$
@FelixMarin Thank you. :-)
Dec
20
answered How find this sum closed form $I=\sum_{k=1}^{n}\int_{0}^{+\infty}\cos{(2kx)}x^{m-1}e^{-ax}dx$