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Can anyone help me with the given integration

Integrate[Exp[- (2  \[Pi]  S  Abs[x-y])/B]* Cos[(2  \[Pi]  S)*(x-y)]* Sin[1 \[Pi] x]Sin[1 \[Pi] y],{x,.001,1}, {y,.001,1}]

Where S and B are input numbers and x and y are variables of integration

3 Replies

I added assumptions, changed the Abs to Sqrt[(x-y)^2], and changed the bottom of the domain from the approximate 0.001 to exact 1/1000 to make things easier for Integrate. This took an hour and a half of CPU time. (The indefinite integral took a couple minutes.)

$Assumptions = {Element[{S,B},Reals], x>0, y>0}
Integrate[(Cos[2*Pi*S*(x-y)]*Sin[Pi*x]*Sin[Pi*y])/E^((2*Pi*S*Sqrt[(x-y)^2])/B),{x,1/1000,1},{y,1/1000,1}]

See the attached notebook, if you really want to. The output is long.

Attachments:
POSTED BY: Bruce Miller
Posted 9 years ago

I stopped Integrate after quite a while -- there may be no closed form for this integral. However, NIntegrate will deliver a number, given values for s and b.

nint[s_, b_] :=
 NIntegrate[
  Exp[-(2 \[Pi] s Abs[x - y])/b]*Cos[(2 \[Pi] s)*(x - y)]*
   Sin[1 \[Pi] x] Sin[1 \[Pi] y], {x, .001, 1}, {y, .001, 1}]

p1 = Plot3D[nint[s, b], {s, 0, 2}, {b, 0, 2}, AxesLabel -> {"S", "B"}]

enter image description here

POSTED BY: David Keith
Posted 9 years ago

I spoke too soon. It does have a symbolic integral. The integral as given does not differ significantly from the one below, in which I changed the lower limits to zero. (Note: it is best to not begin user symbol names with capitals, since they might then conflict with reserved words. )

int0 = Integrate[
  Exp[-(2 \[Pi] S Abs[x - y])/B]*Cos[(2 \[Pi] S)*(x - y)]*
   Sin[1 \[Pi] x] Sin[1 \[Pi] y], {x, 0, 1}, {y, 0, 1}]

(*
(2 \[ExponentialE]^(-((2 \[Pi] S)/B)) (B^8 \[ExponentialE]^((2 \[Pi] \
S)/B)+B^7 \[ExponentialE]^((2 \[Pi] S)/B) \[Pi] S+8 B^6 \
\[ExponentialE]^((2 \[Pi] S)/B) S^2-8 B^8 \[ExponentialE]^((2 \[Pi] \
S)/B) S^2+12 B^5 \[ExponentialE]^((2 \[Pi] S)/B) \[Pi] S^3-4 B^7 \
\[ExponentialE]^((2 \[Pi] S)/B) \[Pi] S^3+16 B^4 \[ExponentialE]^((2 \
\[Pi] S)/B) S^4-96 B^6 \[ExponentialE]^((2 \[Pi] S)/B) S^4+16 B^8 \
\[ExponentialE]^((2 \[Pi] S)/B) S^4+48 B^3 \[ExponentialE]^((2 \[Pi] \
S)/B) \[Pi] S^5+32 B^5 \[ExponentialE]^((2 \[Pi] S)/B) \[Pi] S^5-16 \
B^7 \[ExponentialE]^((2 \[Pi] S)/B) \[Pi] S^5+64 B \
\[ExponentialE]^((2 \[Pi] S)/B) \[Pi] S^7+192 B^3 \[ExponentialE]^((2 \
\[Pi] S)/B) \[Pi] S^7+192 B^5 \[ExponentialE]^((2 \[Pi] S)/B) \[Pi] \
S^7+64 B^7 \[ExponentialE]^((2 \[Pi] S)/B) \[Pi] S^7+B^8 Cos[2 \[Pi] \
S]+8 B^6 S^2 Cos[2 \[Pi] S]-8 B^8 S^2 Cos[2 \[Pi] S]+16 B^4 S^4 Cos[2 \
\[Pi] S]-96 B^6 S^4 Cos[2 \[Pi] S]+16 B^8 S^4 Cos[2 \[Pi] S]-16 B^7 \
S^2 Sin[2 \[Pi] S]-64 B^5 S^4 Sin[2 \[Pi] S]+64 B^7 S^4 Sin[2 \[Pi] \
S]))/(\[Pi]^2 (B^2-4 B^2 S+4 S^2+4 B^2 S^2)^2 (B^2+4 B^2 S+4 S^2+4 \
B^2 S^2)^2)
*)

Plot3D[int0, {S, 0, 2}, {B, 0, 2}, AxesLabel -> {"S", "B"}]

enter image description here

POSTED BY: David Keith
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