Integrals of a rational function multiplied by cos(αx) or sin(αx) use
eiαx=cos(αx)+isin(αx)
and compare real and imaginary parts.
Existence
Let R(z)=Q(z)P(z) where P and Q are polynomials with real coefficients of degrees m and n. Suppose
- Q(x)=0 for all x∈R
- n≥m+1
Then for each α>0, the Cauchy principal values
P.V.∫−∞∞R(x)cos(αx)dxandP.V.∫−∞∞R(x)sin(αx)dx
exist.
Evaluation
Under the same conditions with α>0, let
F(z)=R(z)eiαz
If z1,z2,…,zk are the poles of F in the upper half-plane, then
P.V.∫−∞∞R(x)eiαxdx=2πij=1∑kRes(F,zj)
Consequently,
P.V.∫−∞∞R(x)cos(αx)dx=−2πj=1∑kIm(Res(F,zj))
P.V.∫−∞∞R(x)sin(αx)dx=2πj=1∑kRe(Res(F,zj))
Examples:
- P.V.∫−∞∞x2+4xsinxdx=e2π
- P.V.∫−∞∞x4+4cosxdx=4eπ(cos1+sin1)
Worked Examples
Cosine Integral
Evaluate P.V.∫−∞∞x2+1cosxdx.
Here m=0, n=2≥m+1, α=1, and F(z)=z2+1eiz. The only pole in the upper half-plane is z=i.
Res(F,i)=2iei⋅i=2ie1
Then Res(F,i)=−2ei, so Im(Res(F,i))=−2e1 and Re(Res(F,i))=0.
P.V.∫−∞∞x2+1cosxdx=−2π⋅(−2e1)=eπ
P.V.∫−∞∞x2+1sinxdx=2π⋅0=0
Sine Integral
Show that P.V.∫−∞∞x2+4xsinxdx=e2π.
Here m=1, n=2≥m+1, α=1, and F(z)=z2+4zeiz. The only pole in the upper half-plane is z=2i.
Res(F,2i)=4i2i⋅e−2=2e21
P.V.∫−∞∞x2+4xsinxdx=2πRe(2e21)=e2π
Two Poles
Show that P.V.∫−∞∞x4+4cosxdx=4eπ(cos1+sin1).
Here F(z)=z4+4eiz. The roots of z4=−4 in the upper half-plane are z=1+i and z=−1+i.
For a simple pole zk, using zk4=−4,
Res(F,zk)=4zk3eizk=−16zkeizk
For z=1+i, eiz=e−1ei.
Res(F,1+i)=−16e(1+i)ei
For z=−1+i, eiz=e−1e−i.
Res(F,−1+i)=16e(1−i)e−i
Expand the products.
(1+i)ei=(cos1−sin1)+i(cos1+sin1)
(1−i)e−i=(cos1−sin1)−i(cos1+sin1)
The sum of the residues is
16e−2i(cos1+sin1)=−8ei(cos1+sin1)
P.V.∫−∞∞x4+4cosxdx=−2π(−8ecos1+sin1)=4eπ(cos1+sin1)