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## 6

by: Laurine Willms

54

0

3

# 6 ASE 382R

Laurine Willms
UT
GPA 3.82

Philip Varghese

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COURSE
PROF.
Philip Varghese
TYPE
Class Notes
PAGES
3
WORDS
KARMA
25 ?

## Popular in Aerospace Engineering

This 3 page Class Notes was uploaded by Laurine Willms on Monday September 7, 2015. The Class Notes belongs to ASE 382R at University of Texas at Austin taught by Philip Varghese in Fall. Since its upload, it has received 54 views. For similar materials see /class/181851/ase-382r-university-of-texas-at-austin in Aerospace Engineering at University of Texas at Austin.

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Date Created: 09/07/15
ASE 382R6ME 381Q4 Molecular Gas Dynamics Frequently occu1ing de nite integrals When dealing with distribution functions several integrals appear regularly The integrals tabulated in the back of the text are for limits of 0 and ice When dealing with nite upper or lower bounds a few other intgrals are useful l the error function er x and its complement erfcx and 2 the incomplete gamma lnction F0 06 As noted in Problem 11 51 the error lnction is de ned by er x eXp t2 dt 2 l E 0 The complementary error function is given by erfcx 1 7 er x Note the following er ix 7 er x erf0 0 er w 1 Computer subroutines and even spreadsheet functions are available for computing er x Sketch a graph of equot2 and interpret er x and erfcx graphically The incomplete gamma lnction F0 06 arises when evaluating integrals of the form fvquot eXp Bv2 dv a Such integrals are often referred to as moments of the distribution function It is de ned by F390c Ixj39le39xdx and one can show by direct substitution that 1 1 JVquot eXp Bv2dv n1 Fn2 0 2g 2 3612 A recurrence relation exists for the incomplete gamma function that permits one to reduce j in steps when one wishes to do a numerical calculation F1400j1FJ39 1oc 06H For odd powers of v n odd in the original integral j becomes an integer and one nally has to use a F1a Ix1391e39xdx Ie39xdx f Equivalently using the substitution 3V2 y one could transform the integral to the form I yme ydy where m n71 2 is an integer and then integrate by parts For even n j is halfintegral and the recurrence relation nally requires one to use Hm 17 armE For the special case 060 we get the complete gamma function that satis es the recurrence relation F0 140F041 which gives the following simple results F0 13971 for integer j Fj j lj j lj for half integerj On any test you can leave results directly in terms of 17 06 or erfa On homework problems you should evaluate them numerically to get a feeling for orders of magnitude The following results are useful IexpBV2dV 27212 1 ier a jivexp v2dv 38612 a a it v2 eXp Bv2dv 43933 1i erf E ai aexp a2 239832 no v3 eXp Bv2dv azl 3612 37 v4 eXp Bv2 dv 1i erf F a i aggmg W exP3az avsexp v2dv 23933 22 a2 Bza4 The special case a 0 is dealt with in Appendix 1 of Vincenti and Kruger Occasionally one needs 6 dxE1a x 17005 T where E 1a is the exponential integral a standard integral that is tabulated and available in computer routines era E1a j dx7yilnaizamp 7 057721 56649 is Euler39s constant a quot1 nn x

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