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Anonymous August 11, 2022 (Thu) 13:14:08 No. 9

For γ>0,δ>0

, How do I evaluate this integral?

I=∫H0eitxlog(HH−x)1γ−1⎛⎝(kHlog(HH−x))−1/γ+1⎞⎠−γ−δ−1γ(H−x)dx,

with gratitude.

I tried the usual tricks, with no result so far.

Thank you.

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Anonymous August 09, 2022 (Tue) 23:10:24 No. 13

I need help solving for u here, can anyone help?

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No Anonymous July 31, 2022 (Sun) 12:34:03 No. 10

>>11

Is there a solution for this equation:
x = x!^n
I'm dumb af

Anonymous August 02, 2022 (Tue) 13:37:30 No. 11 >>12

>>10
Yes. You are asking whether $\log x= n\log\Gamma(x+1)$ for some $x$, and presumably some $n$ which is given. It is easy to see that at least for some $n$, such as $n=1,$ this must have (at least) one solution, if you think about the shape of the graphs.

Anonymous August 05, 2022 (Fri) 10:22:01 No. 12

>>11
Oh also, there is of course the trivial solution

x=1

but there is another since the curves have to intersect again.

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Anonymous January 08, 2022 (Sat) 17:27:36 No. 3

>>5 >>8

what did he mean by this? Principles of Mathematics by Carl B page 64, the topic is complex numbers.

Anonymous January 08, 2022 (Sat) 17:37:22 No. 4

Complex numbers are just

\mathbb{R}^2

with an appropriately defined multiplication.

Anonymous February 01, 2022 (Tue) 18:19:45 No. 5

>>3
what do you not understand? the solution for

x^2=-1

Anonymous February 09, 2022 (Wed) 00:30:34 No. 6

When you build a "larger" structure out of a smaller one a copy of the original "smaller" structure is often still present. So complex numbers with no i component are just a copy of real numbers.

Anonymous February 25, 2022 (Fri) 20:21:11 No. 7

It reads like it claims that imaginary numbers enable us to find a solution in term of real numbers.

Anonymous April 17, 2022 (Sun) 17:15:25 No. 8

>>3
>what did he mean by this?
1) All he's doing is writing [math]\mathbb{C} = a + bi \mid a, b \in \mathbb{R}[/math] as 2-tuples [math](a, b)[/math]. The first paragraph is stating something that is otherwise obvious - the subset of complex numbers that have no imaginary part are exactly the real numbers. What separates. [math]\mathbb{C}[/math] from [math]\mathbb{R}^2[/math] with the obvious usual operations entrywise is that imaginary multiplication is different.