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I have a non-analytical equation. I could solve for different values of parameters but my program is not working at all. At the end i want to plot y vs x
f[x_] := y + Sqrt[3 + x*y - x^20 - y^4]
Table[f[x], {x, 0.1, 0.5, 0.1}]
NSolve[f[x] == 0, y]
f[x_] := y + Sqrt[3 + x*y - x^20 - y^4]
sol = Solve[f[x] == 0, y];
x0 = Table[i, {i, 0.1, 0.5, 0.1}];
subs = N[sol /. x -> #] & /# x0
This creates results from which we can see that the first and second solutions produce complex numbers. Plotting the two real solutions first.
y3 = subs[[All, 3, 1, 2]];
y4 = subs[[All, 4, 1, 2]];
ListLinePlot[{Transpose[{x0, y3}], Transpose[{x0, y4}]}]
Alternatively the plot can be produced from the solutions with
Plot[{sol[[3, 1, 2]], sol[[4, 1, 2]]}, {x, 0.1, 0.5}]
The complex solutions can be plotted like so:
ParametricPlot[{{Re[sol[[1, 1, 2]]], Im[sol[[1, 1, 2]]]},
{Re[sol[[2, 1, 2]]], Im[sol[[2, 1, 2]]]}}, {x, 0, Pi/2}]
Somebody have idea how to use all cores for calculating integration? I need to use parallelize or parallel table but how?
f[r_] := Sum[(((-1)^n*(2*r - 2*n - 7)!!)/(2^n*n!*(r - 2*n - 1)!))*
x^(r - 2*n - 1), {n, 0, r/2}];
Nw := Transpose[Table[f[j], {i, 1}, {j, 5, 200, 1}]];
X1 = Integrate[Nw . Transpose[Nw], {x, -1, 1}];
Y1 = Integrate[D[Nw, {x, 2}] . Transpose[D[Nw, {x, 2}]], {x, -1, 1}];
X1//MatrixForm
Y1//MatrixForm
I changed the integration of a list into a list of integrations so that I can use ParallelTable:
X1par=ParallelTable[Integrate[i, {x, -1, 1}], {i, Nw.Transpose[Nw]}];
X1par==X1
(* ===> True *)
Y1par = ParallelTable[Integrate[i,{x,-1,1}],{i,D[Nw,{x,2}].Transpose[D[Nw,{x,2}]]}]
Y1 == Y1par
(* ===> True *)
In my timings, with {j, 5, 30, 1} instead of {j, 5, 200, 1} to restrict the time used somewhat, this is about 3.4 times faster on my quod-core. But it can be done even faster with:
X2par = Parallelize[Integrate[#, {x, -1, 1}] & /# (Nw.Transpose[Nw])]
X2par == X1par == X1
(* ===> True *)
This is about 6.8 times faster, a factor of 2.3 of which is due to Parallelize.
Timing and AbsoluteTiming are not very trustworthy when parallel execution is concerned. I used AbsoluteTime before and after each line and took the difference.
EDIT
We shouldn't forget ParallelMap:
At the coarsest list level (1):
ParallelMap[Integrate[#, {x, -1, 1}] &, Nw.Transpose[Nw], {1}]
At the deepest list level (most fine-grained parallelization):
ParallelMap[Integrate[#, {x, -1, 1}] &, Nw.Transpose[Nw], {2}]
If one helps Integrate a bit by expanding the matrix elements first,
things are doable with a little bit of effort.
On a quad-core laptop with Windows and Mathematica 8.0.4 the following code below runs
for the asked DIM=200 in about 13 minutes,
for DIM=50 the code runs in 6 second.
$starttime = AbsoluteTime[]; Quiet[LaunchKernels[]];
DIM = 200;
Print["$Version = ", $Version, " ||| ", "Number of Kernels : ", Length[Kernels[]]];
f[r_] := f[r] = Sum[(((-1)^n*(-(2*n) + 2*r - 7)!!)*x^(-(2*n) + r - 1))/(2^n*n!*(-(2*n) + r - 1)!), {n, 0, r/2}];
Nw = Transpose[Table[f[j], {i, 1}, {j, 5, DIM, 1}]];
nw2 = Nw . Transpose[Nw];
Print["Seconds for expanding Nw.Transpose[Nm] ", Round[First[AbsoluteTiming[nw3 = ParallelMap[Expand, nw2]; ]]]];
Print["do the integral once: ", Integrate[x^n, {x, -1, 1}, Assumptions -> n > -1]];
Print["the integration can be written as a simple rule: ", intrule = (pol_Plus)?(PolynomialQ[#1, x] & ) :>
(Select[pol, !FreeQ[#1, x] & ] /. x^(n_.) /; n > -1 :> ((-1)^n + 1)/(n + 1)) + 2*(pol /. x -> 0)];
Print["Seconds for integrating Nw.Transpose[Nw] : ", Round[First[AbsoluteTiming[X1 = ParallelTable[row /. intrule, {row, nw3}]; ]]]];
Print["expanding: ", Round[First[AbsoluteTiming[preY1 = ParallelMap[Expand, D[Nw, {x, 2}] . Transpose[D[Nw, {x, 2}]]]; ]]]];
Print["Seconds for integrating : ", Round[First[AbsoluteTiming[Y1 = ParallelTable[py /. intrule, {py, preY1}]; ]]]];
Print["X1 = ", (Shallow[#1, {4, 4}] & )[X1]];
Print["Y1 = ", (Shallow[#1, {4, 4}] & )[Y1]];
Print["seq Y1 : ", Simplify[FindSequenceFunction[Diagonal[Y1], n]]];
Print["seq X1 0 : ",Simplify[FindSequenceFunction[Diagonal[X1, 0], n]]];
Print["seq X1 2: ",Simplify[FindSequenceFunction[Diagonal[X1, 2], n]]];
Print["seq X1 4: ",Simplify[FindSequenceFunction[Diagonal[X1, 4], n]]];
Print["overall time needed in seconds: ", Round[AbsoluteTime[] - $starttime]];
I am wondering if anyone can help me to plot the Cantor dust on the plane in Mathematica. This is linked to the Cantor set.
Thanks a lot.
EDIT
I actually wanted to have something like this:
Here's a naive and probably not very optimized way of reproducing the graphics for the ternary Cantor set construction:
cantorRule = Line[{{a_, n_}, {b_, n_}}] :>
With[{d = b - a, np = n - .1},
{Line[{{a, np}, {a + d/3, np}}], Line[{{b - d/3, np}, {b, np}}]}]
Graphics[{CapForm["Butt"], Thickness[.05],
Flatten#NestList[#/.cantorRule&, Line[{{0., 0}, {1., 0}}], 6]}]
To make Cantor dust using the same replacement rules, we take the result at a particular level, e.g. 4:
dust4=Flatten#Nest[#/.cantorRule&,Line[{{0.,0},{1.,0}}],4]/.Line[{{a_,_},{b_,_}}]:>{a,b}
and take tuples of it
dust4 = Transpose /# Tuples[dust4, 2];
Then we just plot the rectangles
Graphics[Rectangle ### dust4]
Edit: Cantor dust + squares
Changed specs -> New, but similar, solution (still not optimized).
Set n to be a positive integer and choice any subset of 1,...,n then
n = 3; choice = {1, 3};
CanDChoice = c:CanD[__]/;Length[c]===n :> CanD[c[[choice]]];
splitRange = {a_, b_} :> With[{d = (b - a + 0.)/n},
CanD##NestList[# + d &, {a, a + d}, n - 1]];
cantLevToRect[lev_]:=Rectangle###(Transpose/#Tuples[{lev}/.CanD->Sequence,2])
dust = NestList[# /. CanDChoice /. splitRange &, {0, 1}, 4] // Rest;
Graphics[{FaceForm[LightGray], EdgeForm[Black],
Table[cantLevToRect[lev], {lev, Most#dust}],
FaceForm[Black], cantLevToRect[Last#dust /. CanDChoice]}]
Here's the graphics for
n = 7; choice = {1, 2, 4, 6, 7};
dust = NestList[# /. CanDChoice /. splitRange &, {0, 1}, 2] // Rest;
and everything else the same:
Once can use the following approach. Define cantor function:
cantorF[r:(0|1)] = r;
cantorF[r_Rational /; 0 < r < 1] :=
Module[{digs, scale}, {digs, scale} = RealDigits[r, 3];
If[! FreeQ[digs, 1],
digs = Append[TakeWhile[Most[digs]~Join~Last[digs], # != 1 &], 1];];
FromDigits[{digs, scale}, 2]]
Then form the dust by computing differences of F[n/3^k]-F[(n+1/2)/3^k]:
With[{k = 4},
Outer[Times, #, #] &[
Table[(cantorF[(n + 1/2)/3^k] - cantorF[(n)/3^k]), {n, 0,
3^k - 1}]]] // ArrayPlot
I like recursive functions, so
cantor[size_, n_][pt_] :=
With[{s = size/3, ct = cantor[size/3, n - 1]},
{ct[pt], ct[pt + {2 s, 0}], ct[pt + {0, 2 s}], ct[pt + {2 s, 2 s}]}
]
cantor[size_, 0][pt_] := Rectangle[pt, pt + {size, size}]
drawCantor[n_] := Graphics[cantor[1, n][{0, 0}]]
drawCantor[5]
Explanation: size is the edge length of the square the set fits into. pt is the {x,y} coordinates of it lower left corner.
While looking at the belisarius's question about generation of non-singular integer matrices with uniform distribution of its elements, I was studying a paper by Dana Randal, "Efficient generation of random non-singular matrices". The algorithm proposed is recursive, and involves generating a matrix of lower dimension and assigning it to a given minor. I used combinations of Insert and Transpose to do it, but there are must be more efficient ways of doing it. How would you do it?
The following is the code:
Clear[Gen];
Gen[p_, 1] := {{{1}}, RandomInteger[{1, p - 1}, {1, 1}]};
Gen[p_, n_] := Module[{v, r, aa, tt, afr, am, tm},
While[True,
v = RandomInteger[{0, p - 1}, n];
r = LengthWhile[v, # == 0 &] + 1;
If[r <= n, Break[]]
];
afr = UnitVector[n, r];
{am, tm} = Gen[p, n - 1];
{Insert[
Transpose[
Insert[Transpose[am], RandomInteger[{0, p - 1}, n - 1], r]], afr,
1], Insert[
Transpose[Insert[Transpose[tm], ConstantArray[0, n - 1], r]], v,
r]}
]
NonSingularRandomMatrix[p_?PrimeQ, n_] := Mod[Dot ## Gen[p, n], p]
It does generate a non-singular matrix, and has uniform distribution of matrix elements, but requires p to be prime:
The code is also not every efficient, which is, I suspect due to my inefficient matrix constructors:
In[10]:= Timing[NonSingularRandomMatrix[101, 300];]
Out[10]= {0.421, Null}
EDIT So let me condense my question. The minor matrix of a given matrix m can be computed as follows:
MinorMatrix[m_?MatrixQ, {i_, j_}] :=
Drop[Transpose[Drop[Transpose[m], {j}]], {i}]
It is the original matrix with i-th row and j-th column deleted.
I now need to create a matrix of size n by n that will have the given minor matrix mm at position {i,j}. What I used in the algorithm was:
ExpandMinor[minmat_, {i_, j_}, v1_,
v2_] /; {Length[v1] - 1, Length[v2]} == Dimensions[minmat] :=
Insert[Transpose[Insert[Transpose[minmat], v2, j]], v1, i]
Example:
In[31]:= ExpandMinor[
IdentityMatrix[4], {2, 3}, {1, 2, 3, 4, 5}, {2, 3, 4, 4}]
Out[31]= {{1, 0, 2, 0, 0}, {1, 2, 3, 4, 5}, {0, 1, 3, 0, 0}, {0, 0, 4,
1, 0}, {0, 0, 4, 0, 1}}
I am hoping this can be done more efficiently, which is what I am soliciting in the question.
Per blisarius's suggestion I looked into implementing ExpandMinor via ArrayFlatten.
Clear[ExpandMinorAlt];
ExpandMinorAlt[m_, {i_ /; i > 1, j_}, v1_,
v2_] /; {Length[v1] - 1, Length[v2]} == Dimensions[m] :=
ArrayFlatten[{
{Part[m, ;; i - 1, ;; j - 1], Transpose#{v2[[;; i - 1]]},
Part[m, ;; i - 1, j ;;]},
{{v1[[;; j - 1]]}, {{v1[[j]]}}, {v1[[j + 1 ;;]]}},
{Part[m, i ;;, ;; j - 1], Transpose#{v2[[i ;;]]}, Part[m, i ;;, j ;;]}
}]
ExpandMinorAlt[m_, {1, j_}, v1_,
v2_] /; {Length[v1] - 1, Length[v2]} == Dimensions[m] :=
ArrayFlatten[{
{{v1[[;; j - 1]]}, {{v1[[j]]}}, {v1[[j + 1 ;;]]}},
{Part[m, All, ;; j - 1], Transpose#{v2}, Part[m, All, j ;;]}
}]
In[192]:= dim = 5;
mm = RandomInteger[{-5, 5}, {dim, dim}];
v1 = RandomInteger[{-5, 5}, dim + 1];
v2 = RandomInteger[{-5, 5}, dim];
In[196]:=
Table[ExpandMinor[mm, {i, j}, v1, v2] ==
ExpandMinorAlt[mm, {i, j}, v1, v2], {i, dim}, {j, dim}] //
Flatten // DeleteDuplicates
Out[196]= {True}
It took me a while to get here, but since I spent a good part of my postdoc generating random matrices, I could not help it, so here goes. The main inefficiency in the code comes from the necessity to move matrices around (copy them). If we could reformulate the algorithm so that we only modify a single matrix in place, we could win big. For this, we must compute the positions where the inserted vectors/rows will end up, given that we will typically insert in the middle of smaller matrices and thus shift the elements. This is possible. Here is the code:
gen = Compile[{{p, _Integer}, {n, _Integer}},
Module[{vmat = Table[0, {n}, {n}],
rs = Table[0, {n}],(* A vector of r-s*)
amatr = Table[0, {n}, {n}],
tmatr = Table[0, {n}, {n}],
i = 1,
v = Table[0, {n}],
r = n + 1,
rsc = Table[0, {n}], (* recomputed r-s *)
matstarts = Table[0, {n}], (* Horizontal positions of submatrix starts at a given step *)
remainingShifts = Table[0, {n}]
(*
** shifts that will be performed after a given row/vector insertion,
** and can affect the real positions where the elements will end up
*)
},
(*
** Compute the r-s and vectors v all at once. Pad smaller
** vectors v with zeros to fill a rectangular matrix
*)
For[i = 1, i <= n, i++,
While[True,
v = RandomInteger[{0, p - 1}, i];
For[r = 1, r <= i && v[[r]] == 0, r++];
If[r <= i,
vmat[[i]] = PadRight[v, n];
rs[[i]] = r;
Break[]]
]];
(*
** We must recompute the actual r-s, since the elements will
** move due to subsequent column insertions.
** The code below repeatedly adds shifts to the
** r-s on the left, resulting from insertions on the right.
** For example, if vector of r-s
** is {1,2,1,3}, it will become {1,2,1,3}->{2,3,1,3}->{2,4,1,3},
** and the end result shows where
** in the actual matrix the columns (and also rows for the case of
** tmatr) will be inserted
*)
rsc = rs;
For[i = 2, i <= n, i++,
remainingShifts = Take[rsc, i - 1];
For[r = 1, r <= i - 1, r++,
If[remainingShifts[[r]] == rsc[[i]],
Break[]
]
];
If[ r <= n,
rsc[[;; i - 1]] += UnitStep[rsc[[;; i - 1]] - rsc[[i]]]
]
];
(*
** Compute the starting left positions of sub-
** matrices at each step (1x1,2x2,etc)
*)
matstarts = FoldList[Min, First#rsc, Rest#rsc];
(* Initialize matrices - this replaces the recursion base *)
amatr[[n, rsc[[1]]]] = 1;
tmatr[[rsc[[1]], rsc[[1]]]] = RandomInteger[{1, p - 1}];
(* Repeatedly perform insertions - this replaces recursion *)
For[i = 2, i <= n, i++,
amatr[[n - i + 2 ;; n, rsc[[i]]]] = RandomInteger[{0, p - 1}, i - 1];
amatr[[n - i + 1, rsc[[i]]]] = 1;
tmatr[[n - i + 2 ;; n, rsc[[i]]]] = Table[0, {i - 1}];
tmatr[[rsc[[i]],
Fold[# + 1 - Unitize[# - #2] &,
matstarts[[i]] + Range[0, i - 1], Sort[Drop[rsc, i]]]]] =
vmat[[i, 1 ;; i]];
];
{amatr, tmatr}
],
{{FoldList[__], _Integer, 1}}, CompilationTarget -> "C"];
NonSignularRanomMatrix[p_?PrimeQ, n_] := Mod[Dot ## Gen[p, n],p];
NonSignularRanomMatrixAlt[p_?PrimeQ, n_] := Mod[Dot ## gen[p, n],p];
Here is the timing for the large matrix:
In[1114]:= gen [101, 300]; // Timing
Out[1114]= {0.078, Null}
For the histogram, I get the identical plots, and the 10-fold efficiency boost:
In[1118]:=
Histogram[Table[NonSignularRanomMatrix[11, 5][[2, 3]], {10^4}]]; // Timing
Out[1118]= {7.75, Null}
In[1119]:=
Histogram[Table[NonSignularRanomMatrixAlt[11, 5][[2, 3]], {10^4}]]; // Timing
Out[1119]= {0.687, Null}
I expect that upon careful profiling of the above compiled code, one could further improve the performance. Also, I did not use runtime Listable attribute in Compile, while this should be possible. It may also be that the parts of the code which perform assignment to minors are generic enough so that the logic can be factored out of the main function - I did not investigate that yet.
For the first part of your question (which I hope I understand properly) can
MinorMatrix be written as follows?
MinorMatrixAlt[m_?MatrixQ, {i_, j_}] := Drop[mat, {i}, {j}]
What's the best way of getting Mathematica 7 or 8 to do the integral
NIntegrate[Exp[-x]/Sin[Pi x], {x, 0, 50}]
There are poles at every integer - and we want the Cauchy principle value.
The idea is to get a good approximation for the integral from 0 to infinity.
With Integrate there is the option PrincipleValue -> True.
With NIntegrate I can give it the option Exclusions -> (Sin[Pi x] == 0), or manually give it the poles by
NIntegrate[Exp[-x]/Sin[Pi x], Evaluate[{x, 0, Sequence##Range[50], 50}]]
The original command and the above two NIntegrate tricks give the result 60980 +/- 10. But they all spit out errors. What is the best way of getting a quick reliable result for this integral without Mathematica wanting to give errors?
Simon, is there reason to believe your integral is convergent ?
In[52]:= f[k_Integer, eps_Real] :=
NIntegrate[Exp[-x]/Sin[Pi x], {x, k + eps, k + 1 - eps}]
In[53]:= Sum[f[k, 1.0*10^-4], {k, 0, 50}]
Out[53]= 2.72613
In[54]:= Sum[f[k, 1.0*10^-5], {k, 0, 50}]
Out[54]= 3.45906
In[55]:= Sum[f[k, 1.0*10^-6], {k, 0, 50}]
Out[55]= 4.19199
It looks like the problem is at x==0. Splitting integrand k+eps to k+1-eps for integer values of k:
In[65]:= int =
Sum[(-1)^k Exp[-k ], {k, 0, Infinity}] Integrate[
Exp[-x]/Sin[Pi x], {x, eps, 1 - eps}, Assumptions -> 0 < eps < 1/2]
Out[65]= (1/((1 +
E) (I + \[Pi])))E (2 E^(-1 + eps - I eps \[Pi])
Hypergeometric2F1[1, (I + \[Pi])/(2 \[Pi]), 3/2 + I/(2 \[Pi]),
E^(-2 I eps \[Pi])] +
2 E^(I eps (I + \[Pi]))
Hypergeometric2F1[1, (I + \[Pi])/(2 \[Pi]), 3/2 + I/(2 \[Pi]),
E^(2 I eps \[Pi])])
In[73]:= N[int /. eps -> 10^-6, 20]
Out[73]= 4.1919897038160855098 + 0.*10^-20 I
In[74]:= N[int /. eps -> 10^-4, 20]
Out[74]= 2.7261330651934049862 + 0.*10^-20 I
In[75]:= N[int /. eps -> 10^-5, 20]
Out[75]= 3.4590554287709991277 + 0.*10^-20 I
As you see there is a logarithmic singularity.
In[79]:= ser =
Assuming[0 < eps < 1/32, FullSimplify[Series[int, {eps, 0, 1}]]]
Out[79]= SeriesData[eps, 0, {(I*(-1 + E)*Pi -
2*(1 + E)*HarmonicNumber[-(-I + Pi)/(2*Pi)] +
Log[1/(4*eps^2*Pi^2)] - 2*E*Log[2*eps*Pi])/(2*(1 + E)*Pi),
(-1 + E)/((1 + E)*Pi)}, 0, 2, 1]
In[80]:= Normal[
ser] /. {{eps -> 1.*^-6}, {eps -> 0.00001}, {eps -> 0.0001}}
Out[80]= {4.191989703816426 - 7.603403526913691*^-17*I,
3.459055428805136 -
7.603403526913691*^-17*I,
2.726133068607085 - 7.603403526913691*^-17*I}
EDIT
Out[79] of the code above gives the series expansion for eps->0, and if these two logarithmic terms get combined, we get
In[7]:= ser = SeriesData[eps, 0,
{(I*(-1 + E)*Pi - 2*(1 + E)*HarmonicNumber[-(-I + Pi)/(2*Pi)] +
Log[1/(4*eps^2*Pi^2)] - 2*E*Log[2*eps*Pi])/(2*(1 + E)*
Pi),
(-1 + E)/((1 + E)*Pi)}, 0, 2, 1];
In[8]:= Collect[Normal[PowerExpand //# (ser + O[eps])],
Log[eps], FullSimplify]
Out[8]= -(Log[eps]/\[Pi]) + (
I (-1 + E) \[Pi] -
2 (1 + E) (HarmonicNumber[-((-I + \[Pi])/(2 \[Pi]))] +
Log[2 \[Pi]]))/(2 (1 + E) \[Pi])
Clearly the -Log[eps]/Pi came from the pole at x==0. So if one subtracts this, just like principle value method does this for other poles you end up with a finitely value:
In[9]:= % /. Log[eps] -> 0
Out[9]= (I (-1 + E) \[Pi] -
2 (1 + E) (HarmonicNumber[-((-I + \[Pi])/(2 \[Pi]))] +
Log[2 \[Pi]]))/(2 (1 + E) \[Pi])
In[10]:= N[%, 20]
Out[10]= -0.20562403655659928968 + 0.*10^-21 I
Of course, this result is difficult to verify numerically, but you might know more that I do about your problem.
EDIT 2
This edit is to justify In[65] input that computes the original regularized integral. We are computing
Sum[ Integrate[ Exp[-x]/Sin[Pi*x], {x, k+eps, k+1-eps}], {k, 0, Infinity}] ==
Sum[ Integrate[ Exp[-x-k]/Sin[Pi*(k+x)], {x, eps, 1-eps}], {k, 0, Infinity}] ==
Sum[ (-1)^k*Exp[-k]*Integrate[ Exp[-x]/Sin[Pi*x], {x, eps, 1-eps}],
{k, 0, Infinity}] ==
Sum[ (-1)^k*Exp[-k], {k, 0, Infinity}] *
Integrate[ Exp[-x]/Sin[Pi*x], {x, eps, 1-eps}]
In the third line Sin[Pi*(k+x)] == (-1)^k*Sin[Pi*x] for integer k was used.
Simon, I haven't spent much time with your integral, but you should try looking at stationary phase approximation. What you have is a smooth function (exp), and a highly oscillatory function (sine). The work involved is now in brow-beating the 1/sin(x) into the form exp(if(x))
Alternatively, you could use the series expansion of the cosecant (not valid at poles):
In[1]:=Series[Csc[x], {x, 0, 5}]
(formatted) Out[1]=1/x + x/6 + 7/360 x^3 + 31/15120 x^5 +O[x]^6
Note that for all m>-1, you have the following:
In[2]:=Integrate[x^m Exp[-x], {x, 0, Infinity}, Assumptions -> m > -1]
Out[2]=Gamma[1+m]
However, summing the series with the coefficients of cosecant (from wikipedia), not including 1/x Exp[-x] case, which doesn't converge on [0,Infinity].
c[m_] := (-1)^(m + 1) 2 (2^(2 m - 1) - 1) BernoulliB[2 m]/Factorial[2 m];
Sum[c[m] Gamma[1 + 2 m - 1], {m, 1, Infinity}]
does not converge either...
So, I'm not sure that you can work out an approximation for the integral to infinity, but I if you're satisfied with a solution upto some large N, I hope these help.
I have to agree with Sasha, the integral does not appear to be convergent. However, if you exclude x == 0 and break the integral into pieces
Integrate[Exp[-x]/Sin[Pi x], {x, n + 1/2, n + 3/2}, PrincipalValue -> True]
where n >= 0 && Element[n, Integers], then it seems you may get an alternating series
I Sum[ (-1/E)^n, {n, 1, Infinity}] == - I / (1 + E )
Now, I only took it out to n == 4, but it looks reasonable. However, for the integral above with Assumptions -> Element[n, Integers] && n >= 0 Mathematica gives
If[ 2 n >= 1, - I / E, Integrate[ ... ] ]
which just doesn't conform to the individual cases. As an additional note, if the pole lies at the boundary of the integration region, i.e. your limits are {x, n, n + 1}, you only get DirectedInfinitys. A quick look at the plot implies that you with the limits {x, n, n + 1} you only have a strictly positive or negative integrand, so the infinite value may be due to the lack of compensation which {x, n + 1/2, n + 3/2} gives you. Checking with {x, n, n + 2}, however it only spits out the unevaluated integral.