Absolute Value of x from -Pi to Pi.

  part1=Plot[Abs[x+2Pi],{x,-3Pi,-Pi},DisplayFunction->Identity];
  part2=Plot[Abs[x],{x,-Pi,Pi},DisplayFunction->Identity];
  part3=Plot[Abs[x-2Pi],{x,Pi,3Pi},DisplayFunction->Identity];
  absgraph=Show[part1,part2,part3,
  DisplayFunction->$DisplayFunction,AspectRatio->Automatic];

  Clear[a0,a,b,c,Sn]
  a0=(1/(2Pi))(Integrate[-x,{x,-Pi,0}]+Integrate[x,{x,0,Pi}]);
  a0

  a[n_]:=a[n]=(1/Pi)(Integrate[-x Cos[n x],{x,-Pi,0}]+
  Integrate[x Cos[n x],{x,0,Pi}]);
  Table[a[n],{n,1,5}]

  b[n_]:=b[n]=(1/Pi)(Integrate[-x Sin[n x],{x,-Pi,0}]+
  Integrate[x Sin[n x],{x,0,Pi}]);
  Table[b[n],{n,1,5}]

No surprise there, I hope.

  Sn=a0+Sum[a[n]Cos[n x],{n,1,5}]

  p1=Plot[Sn,{x,-Pi,Pi}];

  Show[absgraph,p1];

  Clear[c]

  c[n_]:=c[n]=(1/(2 Pi))(Integrate[-x Exp[- I n x],{x,-Pi,0}]+
  Integrate[x Exp[- I n x],{x,0,Pi}]);
  Table[c[n],{n,-5,5}]

  ISn=Sum[c[n]Exp[I n x],{n,-5,5}]

  ComplexExpand[ISn]

Our partial sums are indeed the same!

  Simplify[Sn-ISn]

  p2=Plot[ISn,{x,-Pi,Pi}];

Up to Examples Using Complex Fourier Series