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4 . M AT H E M AT I C A L N T E R L U D E
In the kinetic theory of gases we deal with integrals of the general type
In([)=5x2n +le- Pdx ([ > 0; n > 1).f we make the substitution, y
=[x2 , the integral reduces to the form
InC[) =!r (n+ l fyne- dy.
However, the factorial function, n ! is dened by
n ! = {Oye- Y dy (n > 1)
(4.35)
(4.36)
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so that
InC[)= Lon+1e- Px2dx=![- (n +1 ). (4. 37)The higher-order integrals can be obtained from those of lower order by dierentiatin ;dierentiating Eq. (4.37) with respect to [ yields
or
dU[)
d[
Two cases commonly arise.
Case I. n = or a positie integer.
(4.38)
In this case we apply Eq . (4.37) directly and no iculty ensues . The lowest member is
0([) = p-
.
All other members can be obtained from Eq. (4.37) or by dierentiating 0([) and using
Eq. (4.38).
Case n. n =
-1, 1, t orn = m
- 1 whrem =
or a positie integer.In this case we may also use Eq. (4.37) directly, but unless we know the alue of the faorialfunction for half-integral alues of the argument we will be in trouble . If n = m thefunction takes the form
Im- 1 /2([)= Lome - Px2 dx=H(m-)!][(m+1/2) . (4.39)When m = 0, w have
L/2([) =
Loepx2 dx = [- /2
LOe-
y2 dy = [- /2L /z (1), (4 .40)
where in the second writing, x = [- /2y, has been used. Comparing this result with the
last member of Eq. (4.39) we nd that
L /2(1) = Loe y2 dy = (1) ! . (4.41)The integral, - /2(1), cannot be ealuated by elementary methods. We proceed by writingthe integral in two ways,
_ /2(1)=
LOex2 dxthen multiply them together to obtain,
and
/2(1) = LoLe- (x2 + y2) dxdy.The integration is oer the area of the rst quadrat; we change ariables to r2 = x2 + y2
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and replace dx dy by the element of area in polar coordinates, r d dr. To cover the rstquadrant we integrate from zero to l and r from 0 to 0 : the integral becomes
I 1 /2( ) = f/2d Ie- 2r dr= G) 1e- 2 d(r2)= Ie - Y dy.The last integral is equal to O = ; taking the square root of both sides, e have
I - l i) = !I (44)Comparing Eqs (4.41) and (4.4), it follows that (! = I; now from Eqs. (4.40) and(4.4),
L1/2(P) = !Jp- 1/2 By dierentiation, and by using Eq (4.38) we obtain
and
I (P) dL 1/2 _ 1p- 3/2)1 /2 2y 2dP
I (P) = _ dI1 /2 = .1p 5/2)3/2 dP 2y n 2 2 .
Repetition of this procedure ultimately yields
I =foxme- PX2dx= (2m) .!p- (m + 1/)m- 1 /2 2y 2m o m . (4.43)By comparing this result with Eq (4.39) we obtain the interesting reslt fo half-integralfactorials
( _ ) 1 = (m) !m 2 y 2m , .m .Table 4. collects the most commonly used formulas
Table 4 . 1
ntegrals tat occur i n te k inetic teory of gases
(3) x2e- p dx = tftP- 3/2
(4) x4e- dx=tfiP- S/2
fO n !-(+ l/2 )(5) x2ne - P dx=1fo 2 n
(6) f x2+ le- dx = O0
(7) foxe - dx = tP- 1
(8) 1x3e- dx = tP-
(9) 10xSe- dx = - 3
(4.44)
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* 47 . 1 The Errr Fu ct ion
We frequently have occasion to use integrals o f the type o f Case above in which the
upper limit is not extended to innity but only to some nite vlue These integrals arerelated to the error function (erf) We den
erf () = e- U2 duo (445)f the upper limit is extended to - 0, the integral is t. so that
erf() = 1
Thus as varies from zero to innity, erf () varies from zero t o unity fwe add the integral
from to 0 multiplied by 2/. to both sides of the equation, we obtain2 fo 2 [f" fo ] 2 fo
erf () + " e- u2 du =. 0 e- u2 du + " e- U2 du = 0 e-u2 du = 1 Therefore
2 fo. "
e-u2 du = 1 - erf ()
We dene the co-error function, erfc (), by
erfc () = 1 - erf ()
Thus
Some values of the error function are given in Table 42
x erf(x)
0.00 0.000
0.10 0 . 1 12
0.20 0.2230.30 0.329
0.40 0.428
0.50 0. 5210.60 0.604
0.70 0.678
Table 4 .2
The error function :
2
IX
erf(x) = I 0 e- , 2 du
x erf(x)
0.80 0.742
0.90 0.797
1.00 0.8431.10 0.880
1.20 0.910
1.30 0.934
1.40 0.952
1.50 0.966
x
1.60
1.70
1.801.90
2.00
2.20
2.40
2.50
(446)
(447)
erf(x)
0.976
0.984
0.9890.993
0.995
0.998
0.9993
0.9996
