Problem integral pi over e
Problem integral pi over e
Prove that
In[]:=
Integrate[x*Sin[x]/(1+x^2),{x,0,Infinity}]
Out[]=
π
2
Solution (José Manuel Rodríguez Caballero)
From the identity
In[]:=
FullSimplify[1/z+2*z*Sum[1/(z^2-n^2),{n,1,Infinity}]]
Out[]=
πCot[πz]
which is the result of the logarithmic derivative applied to the product defining the sinus, we obtain
In[]:=
x/(1+x^2)+Sum[(x+2*n*Pi)/(1+(x+2*n*Pi)^2)+(x-2*n*Pi)/(1+(x-2*n*Pi)^2),{n,1,Infinity}]//FullSimplify
Out[]=
1
4
1
2
1
2
Multiplying by Sin[x] and Integrating from 0 to 2π, on one side we obtain
In[]:=
(-Integrate[Sin[x]*Cot[(-I-x)/2],{x,0,2*Pi}]+Integrate[Sin[x]*Cot[(-I+x)/2],{x,0,2*Pi}])/4
Out[]=
π
on the other side we have
Integrate[ Sin[x] * (x/(1 + x^2) + Sum[(x + 2*n*Pi)/(1 + (x + 2*n*Pi)^2) + (x - 2*n*Pi)/(1 + (x - 2*n*Pi)^2), {n, 1, Infinity}], {x, 0, 2*Pi})]
= Integrate[ Sin[x] * Sum[(x + 2*n*Pi)/(1 + (x + 2*n*Pi)^2)), {n, -Infinity, Infinity}], {x, 0, 2*Pi}]
= Integrate[ Sum[Sin[x] * (x + 2*n*Pi)/(1 + (x + 2*n*Pi)^2)), {n, -Infinity, Infinity}], {x, 0, 2*Pi}]
= Integrate[ Sum[Sin[x + 2*n*Pi] * (x + 2*n*Pi)/(1 + (x + 2*n*Pi)^2)), {n, -Infinity, Infinity}], {x, 0, 2*Pi}]
= Sum[ Integrate[ Sin[x + 2*n*Pi] * (x + 2*n*Pi)/(1 + (x + 2*n*Pi)^2)), {x, 0, 2*Pi}], {n, -Infinity, Infinity}]
= Sum[ Integrate[ Sin[x] * x/(1 + x^2)), {x, 2*n*Pi, 2*(n+1)*Pi}], {n, -Infinity, Infinity}]
= Integrate[ Sin[x] * x/(1 + x^2)), {x, -Infinity, Infinity}]
= 2 * Integrate[ Sin[x] * x/(1 + x^2)), {x, 0, Infinity}]
Therefore,
Integrate[ Sin[x] * x/(1 + x^2)), {x, 0, Infinity}] = Pi/(2*E)
Integrate[ Sin[x] * x/(1 + x^2)), {x, 0, Infinity}] = Pi/(2*E)