stakgold chap 7 exercise 7_54
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Short worked solution dated 12.31.11, written by Phil as part of his Stakgold Chapter 7 support files. He defines F(a) as an integral of e^{-a(x+p)}/[(x+p)(x-k)^{1/2}], differentiates with respect to a, and reduces the result to a Gamma-type integral worth sqrt(pi). Integrating back gives an error function, and the condition F(infinity)=0 fixes the constant, so F(a) = (pi/sqrt(c)) erfc(sqrt(c a)) with c=p+k, matching Stakgold's answer.
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Exercise 7.54 PhL 12.31.11
Do a certain integral
F(a) = !Syntax Error, Idx e-a(x+p) / [ (x+p)(x-k)1/2 ]
dF/da = !Syntax Error, Idx (-1)(x+p)e-a(x+p) / [ (x+p)(x-k)1/2 ]
= - !Syntax Error, Idx e-a(x+p)/ (x-k)1/2
= - e-ap !Syntax Error, Idx e-ax/ (x-k)1/2
Change to y = x-k so dy = dx to get
= - e-ap !Syntax Error, Idy e-a(y+k)/ y1/2
= - e-ap e-ak !Syntax Error, Idy e-ay / y1/2
Change to z = ay so dz = ady
= -e-a(p+k) (1/a) !Syntax Error, Idz e-z / (z/a)1/2
= -e-a(p+k) a-1 a1/2 !Syntax Error, Idz e-z/z1/2
The integral is π1/2 says Maple, so
dF/da = - π1/2 a-1/2 e-a(p+k)
= - π1/2 (e-a(p+k)/ )
dF = - π1/2 (e-a(p+k)/ ) da
Now what? Integrate both sides to get
F(a) = - π1/2 !Syntax Error, Ida' (e-a'(p+k)/ ) + constant
Let p+k = c so this says
F(a) = - π1/2 !Syntax Error, Idx (e-cx)/ ) + constant
Maple says the integral is
so we get
F(a) = - π1/2 π1/2 erf(c1/2 a1/2) / c1/2 + constant
Now if a = 0 we have erf(0) which is 0 so we get
F(0) = 0 + constant =>
F(a) = - π erf(c1/2 a1/2) / c1/2 + F(0)
We also know that erf(∞) = 1 so
F(∞) = - π/c1/2 + F(0)
Now look at our original integral
F(a) = !Syntax Error, Idx e-a(x+p) / [ (x+p)(x-k)1/2 ]
It sure seems that F(∞) = 0 here, so then
F(0) = π/c1/2
and then
F(a) = - π erf(c1/2 a1/2) / c1/2 + π/c1/2
= π/c1/2 [ 1 - erf(c1/2 a1/2)] = π/c1/2 erfc(c1/2 a1/2)
= ( π/) erfc( )
and we then get Stak's result ! There must be an easier way to do this integral but OK.