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

zakinutuzi 2021-12-30 Answered
Solve integrals using residue theorem?
0πdθ2+cosθ
0x(1+x)6dx
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kalfswors0m
Answered 2021-12-31 Author has 24 answers
Hint for the First one First compute
I=02πdθ2+cosθ=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+z12
sin(t)=Im(z)=zz12i
dz=ieitdt=izdtdt=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,z1z2i)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+cosθ=I2
and hence your result follows by computing the next integral
0πdθ2+cosθ=1i|z|=1dzz2+4z+1
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Linda Birchfield
Answered 2022-01-01 Author has 39 answers
For the second integral:
Note first that this integral is easily done by recognizing that x=(1+x)1, so the integral is really
0dx(1+x)50dx(1+x)6=1415=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
Cdzzlogz(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
ϵRdxxlogx(1+x)6+iR02πdθeiθReiθlog(Reiθ}{(1+Reiθ)6}
+Rϵdxx(logx+i2π)(1+x)6+iϵ2π0dϕ
eiϕϵeiϕlog(ϵeiϕ)(1+ϵeiϕ)6
As R, the second integral vanishes as logRR4. As ϵ0, the fourth integral vanishes as ϵ2logϵ. Thus, the contour integral is, in this limit
i2π0dxx(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(zlogz)]z=eiπ=3!5!
Putting this altogether, we get that
0dxx(1+x)6=120
which agrees with the above.
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karton
Answered 2022-01-09 Author has 368 answers

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(m1)!limzz0dm1dzm1((zz0)mf(z)).
And finally, apply the Residue Theorem.

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