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Nilknarf
12:13 AM
@SimplyBeautifulArt Aw, that's a dirty trick. You gave me an essential singularity. D:<
Hahaha, I got it
$$\huge{2\pi}$$
@SimplyBeautifulArt Where do you find problems like that?
20 hours later…
Simply Beautiful Art
8:02 PM
@Nilknarf I happened to find that one on math.se
@Nilknarf Oh dear
Nilknarf
8:30 PM
Hey!
@SimplyBeautifulArt Got anymore like that?
Simply Beautiful Art
Oh hey! @Nilknarf
Wow, I'm not late
Nilknarf
Also, I just realized that it can be done without worrying about essential singularities
Yay, you're here
Simply Beautiful Art
Yeah, residue theorem doesn't care about whether or not it's essential
Nilknarf
How do you like my nasty dottie integral? :D
Simply Beautiful Art
I was (and still am) playing English shiritori
And yes, I like it
And I'm also explaining S.O.A.P. to math.se user Wythagoras.
Nilknarf
8:33 PM
Ooh
Do you have time to help me w/ an integral?
Simply Beautiful Art
maybe
Nilknarf
$$\Gamma'(1)$$
Simply Beautiful Art
Depends on when I'm leaving, which isn't happening while my mom is on the phone lol
Nilknarf
Ah
Simply Beautiful Art
So go on
Nilknarf
8:35 PM
$$\int_0^\infty e^{-x}\ln(x)dx$$
$$=-\gamma$$
But I can't figure out how to prove it.
Maybe I have to split it up or something
Leaky Nun
$\displaystyle \Gamma(t) := \int_0^\infty x^{t-1} e^{-x} \ \mathrm dx$
Nilknarf
$$\int_0^1 e^{-x}\ln(x)dx+\int_1^\infty e^{-x}\ln(x)dx$$
Simply Beautiful Art
Hey @LeakyNun
Leaky Nun
hi
Simply Beautiful Art
What's your definition of $\gamma$?
Nilknarf
8:40 PM
$$\gamma = \int_1^\infty \bigg(\frac{1}{\lfloor x\rfloor}-\frac{1}{x}\bigg)dx$$
Leaky Nun
$\displaystyle \Gamma'(t) := \int_0^\infty \frac {\partial} {\partial t} x^{t-1} e^{-x} \ \mathrm dx$
$\displaystyle \Gamma'(t) := \int_0^\infty x^{t-1} e^{-x} \ln x \ \mathrm dx$
$\displaystyle \Gamma'(1) := \int_0^\infty e^{-x} \ln x \ \mathrm dx$
Nilknarf
Right
Simply Beautiful Art
Hey @Mr.Xcoder
Have you checked Discord?
Mr. Xcoder
Hey @SBA
Not in while
Simply Beautiful Art
U should
And it's time for me to go
Gah
:'(
Nilknarf
8:43 PM
:P
Mr. Xcoder
@SimplyBeautifulArt Will do. sya
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