Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

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## #1 2013-05-07 01:02:09

ShivamS
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### Some contour integration

99% chance I am wrong...
Also, do not know Latex.
I =  ∫  infinity on top, 0 on bottom     (x^alpha)/(1+ √2x +x^2)dx

I am getting I = (√2pi)sin(alpha pi/4)  /  sin(alpha pi)

I have discovered a truly marvellous signature, which this margin is too narrow to contain. -Fermat
Give me a lever long enough and a fulcrum on which to place it, and I shall move the world. -Archimedes
Young man, in mathematics you don't understand things. You just get used to them. - Neumann

## #2 2013-05-07 02:20:52

anonimnystefy
Real Member

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### Re: Some contour integration

M is getting

for -1<alpha<1.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #3 2013-05-08 22:47:16

ShivamS
Super Member

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### Re: Some contour integration

How do you get that? And M?

I have discovered a truly marvellous signature, which this margin is too narrow to contain. -Fermat
Give me a lever long enough and a fulcrum on which to place it, and I shall move the world. -Archimedes
Young man, in mathematics you don't understand things. You just get used to them. - Neumann

## #4 2013-05-09 03:08:49

anonimnystefy
Real Member

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### Re: Some contour integration

Mathematica.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment