Solve integrals using residue theorem? ∫0πdθ2+cos⁡θ ∫0∞x(1+x)6dx

Question

Solve integrals using residue theorem?  ∫0πdθ2+cos⁡θ  ∫0∞x(1+x)6dx

kalfswors0m · Accepted Answer

Hint for the First one First compute  I=∫02πdθ2+cos⁡θ=∫02πR(cos⁡(θ), sin⁡(θ))dθ  Where R is the rational function given by  R(x,y)=12+x  How to do this using the residue theorem? Put z=eit=cos⁡(t)+isin⁡(t), thus  cos⁡(t)=Re(z)=z+z−12  sin⁡(t)=Im(z)=z−z−12i  dz=ieitdt=izdt⇒dt=1izdz  Then I can be seen as a contour integral, solve it by using residues  I=∫02πR(cos⁡(t),sin⁡(t))dt=  ∫|z|=1R(z+1z2,z−1z2i)1izdz  Hence in your case the integral you will compute is  I=∫|z|=1(12+z+1z2)1izdz=2i  ∫|z|=1dzz2+4z+1  which can be easily obtain by the residue theorem!  Finally: Note that  ∫0πdθ2+cos⁡θ=I2  and hence your result follows by computing the next integral  ∫0πdθ2+cos⁡θ=1i∫|z|=1dzz2+4z+1

Linda Birchfield · Accepted Answer

For the second integral:  Note first that this integral is easily done by recognizing that x=(1+x)−1, so the integral is really  ∫0∞dx(1+x)5−∫0∞dx(1+x)6=14−15=120  One may also use the residue theorem. However, one must choose an appropriate contour and integrand. In this case, a useful contour integral to consider is  ∮Cdzzlog⁡z(1+z)6  where C is a keyhole contour of outer radius R about the positive real axis. The contour integral is then equal to  ∫ϵRdxxlog⁡x(1+x)6+iR∫02πdθeiθReiθlog⁡(Reiθ}{(1+Reiθ)6}  +∫Rϵdxx(log⁡x+i2π)(1+x)6+iϵ∫2π0dϕ  eiϕϵeiϕlog⁡(ϵeiϕ)(1+ϵeiϕ)6  As R→∞, the second integral vanishes as log⁡RR4. As ϵ→0, the fourth integral vanishes as  ϵ2log⁡ϵ. Thus, the contour integral is, in this limit  −i2π∫0∞dxx(1+x)6  By the residue theorem, the contour integral is also equal to i2π times the residue at the pole x=eiπ. (Note how important it is to get the argument correct.) The residue at this pole is  15![d5dz5(zlog⁡z)]z=eiπ=−3!5!  Putting this altogether, we get that  ∫0∞dxx(1+x)6=120  which agrees with the above.

karton · Accepted Answer

Hint for the first one: Consider the function f(z)=2z2+4z+1 and find its poles. Then use the known formula for residues: Under the assumption that f has a pole of order m at z0 Res(z0,f)=1(m−1)!limz→z0dm−1dzm−1((z−z0)mf(z)). And finally, apply the Residue Theorem.

Solve integrals using residue theorem? \int_0^{\pi}\frac{d\theta}{2+\cos\theta} \int_0^\infty\frac{x}{(1+x)^6}dx

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