## Empty one year ago Quick little thing explaining how to get the gamma function integral @astrophysics

1. Empty

Ok solve this integral, a is a constant with respect to x. $$\int_0^\infty e^{-ax}dx$$

2. Astrophysics

Wait wait do this

3. Astrophysics

|dw:1437992640744:dw| i saw this and wanted to do it

4. Empty

Yeah I know, trust me

5. Empty

lol

6. Astrophysics

by parts

7. Astrophysics

ok

8. Astrophysics

ima do it then

9. Empty

By parts is one way to do it, but this way is much cooler

10. Empty

Hahaha trust me you will be doing the same thing but cooler

11. Empty

don't you wanna be cool

12. Empty

all the kids are doing it

13. Empty

it's just one integral

14. Empty

one can't hurt

15. Empty

Woah this is OPEN STUDY not PEER PRESSURE ANSWER

16. Empty

c'mon what's the integral of e^-ax already hurrrrry

17. Empty

ty

18. Astrophysics

|dw:1437993094157:dw| I forgot to take the limit the first time ok, relax it's been a while since I've done an improper integral

19. Empty

haha yeah no problem this looks perfect!

20. Astrophysics

Jk I did one earlier...|dw:1437993189968:dw|

21. Astrophysics

|dw:1437993463484:dw|

22. Empty

Boom ok perfect now we have this nice looking thing: $$a^{-1} = \int_0^\infty e^{-ax}dx$$ This is where the magic happens, this is the cool way. We differentiate both sides WITH RESPECT TO A. This is weird if you're not used to it, so it'll be good practice. If you don't make a mistake doing this, then you will be officially better than me cause I messed this up the first time I saw it lol.

23. Astrophysics

I am better

24. Empty

lolol woahhhh champ

25. Empty

I was gonna give you a medal + fan but now I'm considering giving it to Jhan

26. ParthKohli

JUST DO IT.

27. anonymous

|dw:1437993589194:dw|

28. Astrophysics

I take that back, please I need my medals and fans ok leggo

29. Empty

lolol

30. Astrophysics

I tried putting the a over the t JB

31. Empty

Parth gets the medal

32. Empty

sry

33. Empty

I haven't got all night

34. ParthKohli

It's because I'm better than him.

35. Empty

I'm gonna pull a Michele_Laino on you if you don't hurry up

36. Astrophysics

Go do your discriminants and quit discriminating against me so I can do this

37. Astrophysics

|dw:1437993707615:dw|

38. Astrophysics

Differentiate both sides you say?

39. Empty

yeah with respect to a

40. Empty

and with respect to me, your sensei

41. Astrophysics

|dw:1437993761110:dw| I'm better

42. ParthKohli

FLAWLESS VICTORY

43. anonymous

Why take the limit when you can differentiate - his logic.

44. Empty

|dw:1437994060905:dw| You evaluated the left side, but not the right side and now Parth is enjoying his birthday wish pssssh

45. ParthKohli

Yes, my birthday wish was a free medal. <3

46. Astrophysics

That was the right side

47. Astrophysics

|dw:1437994295728:dw|

48. Empty

$$\frac{d}{da} \int_0^\infty e^{-ax}dx = \int_0^\infty \frac{\partial}{\partial a} (e^{-ax})dx$$

49. Astrophysics

Yeah ok I see what you're doing, you want me to do use leibniz rule pretty much and not just plug and chug haha

50. anonymous

defining a is indeed a constant and not just another variable

51. Empty

a is not a constant, it is just an independent variable from x

52. Astrophysics

|dw:1437994507765:dw| oOoOoOo

53. Empty

Constant with respect to x though, you could say that, also https://www.youtube.com/watch?v=Lcw6xBaCaXM

54. Empty

Boom good job, ok now we have something good

55. anonymous

Ah, I see.

56. Empty

$a^{-1} = \int_0^\infty e^{-ax}dx$ $a^{-2} = \int_0^\infty xe^{-ax}dx$ ...? Keep going do this 2 or 3 more times until you get the picture.

57. Empty

differentiate with respect to a I mean haha

58. Astrophysics

Integrate it, or keep taking derivatives?

59. Empty

keep taking derivatives :D

60. Astrophysics

K good lol

61. Astrophysics

|dw:1437994844847:dw|

62. Empty

noice

63. Astrophysics

I think we can write a series for this...

64. Empty

no don't!

65. Empty

just keep taking derivatives

66. anonymous

how do partials work again? $$a^{-2} = \int_0^\infty \color{red}{x}e^{-ax}dx$$ work again? The highlighted red portion I mean...

67. Astrophysics

Ok I'll take one more lol

68. Empty

x is a constant when we're differentiating with respect to a @Jhannybean since a and x are independent of each other nothing happens

69. anonymous

Oh that's right,

70. anonymous

I kept reading $$\frac{ \partial}{\partial x}$$ instead of $$\frac{\partial}{\partial a}$$ and I was confusing myself. -_-

71. Astrophysics

Taking derivatives in integrals is amazing, I saw this technique once before, did not know you would treat it like a constant in the integral, awesome. |dw:1437995191725:dw|

72. Astrophysics

Oki, what's next

73. Empty

Do one more!

74. Astrophysics

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75. Astrophysics

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76. Empty

$a^{-1} = \int_0^\infty e^{-ax}dx$ $a^{-2} = \int_0^\infty xe^{-ax}dx$ $6a^{-3} = \int_0^\infty x^2e^{-ax}dx$ $24a^{-4} = \int_0^\infty x^3e^{-ax}dx$ ...?

77. Empty

What are we doing again

78. Empty

something something gamma is the continuous version of the factorial something something whatever I already forgot

79. Astrophysics

Ok I see, there's a factorial relationship, so there's a reoccurrence relationship.

80. Empty

Reoccurrence relationship?

81. Empty

$a^{-1} = \int_0^\infty e^{-ax}dx$ $a^{-2} = \int_0^\infty xe^{-ax}dx$ $6a^{-3} = \int_0^\infty x^2e^{-ax}dx$ $24a^{-4} = \int_0^\infty x^3e^{-ax}dx$ ...? This is the pattern, what's the general form?

82. Astrophysics

Haha recurrence*

83. Empty

No recurrence relation here either

84. Empty

Whoops I had been writing it wrong lol oh well don't look at what I write you wrote it yourself correctly anyways $a^{-1} = \int_0^\infty e^{-ax}dx$ $a^{-2} = \int_0^\infty xe^{-ax}dx$ $2a^{-3} = \int_0^\infty x^2e^{-ax}dx$ $6a^{-4} = \int_0^\infty x^3e^{-ax}dx$ $24a^{-5} = \int_0^\infty x^4e^{-ax}dx$ ... $\frac{n!}{a^{-???} }= \int_0^\infty ???dx$ Here's my hint, what the hell is this last thing!

85. Astrophysics

(n-1)!

86. Empty

Stop looking at what it should be and look at what it actually is, forget the gamma function for a minute

87. Empty

Just find the truth don't worry about being right right now

88. Empty

idk check it: Here's what you KNOW $2a^{-3} = \int_0^\infty x^2e^{-ax}dx$ try out your false formula to find out how to fix it

89. Astrophysics

Yeah I tried, that didn't work..one sec let me go over it again..6 am xD

90. Empty

Pythagoras didn't have anyone to check his work I think on the whole a^2+b^2 thing lol so you're at least fortunate :P

91. Astrophysics

$\frac{ n! }{ a^{-(n+1)} } = \int\limits_{0}^{\infty} x^{n+1} e^{-ax} dx$

92. Astrophysics

I tried by plugging in integers haha

93. Empty

Known: $\frac{2!}{a^3}= \int_0^\infty x^2e^{-ax}dx$ Yours: $\frac{ n! }{ a^{-(n+1)} } = \int\limits_{0}^{\infty} x^{n+1} e^{-ax} dx$ Plugged in n=2 $\frac{ 2! }{ a^{-3} } = \int\limits_{0}^{\infty} x^{3} e^{-ax} dx$ Same or not?

94. Astrophysics

x^n but the known should be a^-(...) ye?

95. Empty

???

96. Empty

Sorry I don't know what you're saying X_X

97. Astrophysics

Oh no I was just looking at when you said $\frac{n!}{a^{-???} }= \int\limits_0^\infty ???dx$

98. Astrophysics

$\frac{ n! }{ a^{-(n+1)} } = \int\limits_{0}^{\infty} x^n e^{-ax} dx$

99. Empty

better but why are you just going with my mistake?

100. Empty

like negative in the denominator is like "double dividing"

101. Astrophysics

Haha, I get scared so I thought it was right

102. Astrophysics

That works out

103. Astrophysics

Tried n = 2

104. anonymous

My gawd this is a long question

105. Empty

Hahaha ok that's what I like to hear, everything leading to this point doesn't really matter now, it's like the ladder we can kick out from under us and if you like you can pretend we were never wrong on the path to this. @isaac4321 yeah we should probably have flushed halfway through for courtesy :(

106. anonymous

It's all good at least your gettin edumacated

107. Empty

tru

108. Astrophysics

|dw:1437998982910:dw|

109. anonymous

Um ok

110. Astrophysics

Wow, empty, this happens every time spend so much time on one thing..then...yeah it doesn't matter xD

111. Astrophysics

You're like a real prof

112. anonymous

Who is that

113. Astrophysics

114. anonymous

Ok I thought it was him

115. Empty

ok so now you can derive the gamma function, you're on your own, you can think more about why this works

116. anonymous

👍

117. Empty

It should be really easy

118. Empty

and if it isn't, wellllll think about it until you do

119. Astrophysics

Ok so I don't really know what exactly the gamma function is, I mean is it actually the thing I posted above, I thought that was just a problem that was a gamma function haha..well that's cool, learned so much

120. anonymous

Obviously it is very easy 😉

121. Astrophysics

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122. Empty

Yeah, the gamma function is basically defined as $\Gamma(n)=(n-1)!$ for positive integers but extends further.

123. Empty

do you see where the n-1 is coming into play now?

124. Astrophysics

Yeah I think so

125. Empty

ok

126. Empty

|dw:1437999792138:dw|

127. Astrophysics

lmao

128. Astrophysics

Howd you get so many medals

129. Empty

it's a thumbs up not a pile of poop for the record

130. Astrophysics

Why do people always observe you teaching me

131. Empty

they're creepers

132. Empty

creepin'

133. Empty

idk because this site is boring normally lo

134. Astrophysics

So I don't really see where exactly to use this function, like would you use it in QM, looks like it may have to do with statistics/ probability.

135. Empty

Yeah I have used it in QM which is a statistical theory, so yes to both. You can use it anywhere you would use a factorial.