Principal value integral

(FB/Math/BM – Integral contributed by Sourav Hom Choudhury)

Integrate ∫ 1/[(x²+4)²(x+1)] dx from x = -∞ to x = ∞.

____________________________________________________

That integral is divergent, because of the simple pole at z = -1.

I calculated the integral, interpreted as principal value, with two methods (must for fun but I used Derive for dull routines).

I. Using partial fractions:

We first find a primitive funtion (aka antiderivative aka indeterminate integral):

J(x) ≡ ∫ 1/(x² + 4)² (x + 1) dx =

= ∫ (1 – x)/(5·(x² + 4)²) + (1 – x)/(25·(x² + 4)) + 1/(25·(x + 1)) dx =

= (13/400)·arctan(x/2) – (1/50)·ln(x² + 4) + (1/25)·ln |x + 1|
+ (1/40)·(x + 4)/(x² + 4)

Then by the definition of principal value (see previous post),

P ∫ [-∞,∞] 1/(x² + 4)² (x + 1) dx =

= lim [ε→0+] { lim [x→∞] J(x) – J(-1+ε) + J(-1-ε) – lim [x→-∞] J(x) } =

= lim [ε→0+] { 13π/800 – J(-1+ε) + J(-1-ε) – (-13π/800) } =

= 13π/400 + lim [ε→0+] { – J(-1+ε) + J(-1-ε) }. Now,

– J(-1+ε) + J(-1-ε) =

= -(13/400)·arctan((e – 1)/2) – (13/400)·arctan((e + 1)/2)
+ (1/50)·ln((ε² – 2ε + 5)/(ε² + 2ε + 5))
– (1/20)·ε(ε^2 + 11)/((ε² + 2ε + 5)·(ε² – 2ε + 5))

and from this

lim [ε→0+] { – J(-1+ε) + J(-1-ε) } =
= -(13/400)·arctan(-1/2) – (13/400)·arctan(1/2) + (1/50)·ln 1- (1/20)·0
= 0

so the integral is

P ∫ [-∞,∞] 1/(x² + 4)² (x + 1) dx = 13π/400.

It should be noted that the separat limits lim [ε→0+] J(-1+ε) and [ε→0+] J(-1-ε) do not exist, which is why the integral in the usual sense is divergent.
________________________________
2. By residue calculus:

There is a double pole at z = 2i with residue
lim [z→2i] d/dz { (z – 2i)² / ( (z² + 4)²·(z + 1) ) } =
= lim [z→2i] d/dz { 1 / ( (z + 2i)²·(z + 1) ) } = -1/50 – 13i/800

and a simple pole at z = -1 with residue
lim [z→-1] { 1/(z² + 4)² } = 1/25

so the integral is

P ∫ [-∞,∞] 1/(x² + 4)² (x + 1) dx =

= 2πi*(-1/50 – 13i/800) + πi*(1/25) = 2π*13/800 = 13π/400.
——————————————–
The reason that the pole at z = -1 only gives half the residue:

We start with a integration path consisting of the real interval [-R,-1-ε], a half-circle in negative sense (clockwise) and centre -1 , the real interval [-1+ε,R], and a half-circle in in positive direction (counterclockwise). Then, by the residue theorem,

(*) ∫ [-R,-1-ε] f(x) dx + ∫ [|z+1|=ε] f(z) dz + ∫ [-1+ε,R] f(x) dx
+ ∫ [|z|=R] f(z) dz = 2πi*Res(z=2i, f(z))

∫ [|z|=a] f(z) dz = πi*Res(z=0), integrated over a half-circle in positive sense, since only the 1/z-term in the Laurent expansion contributes. Also ∫ [|z|=R] f(z) dz → 0 as R → ∞, since |f(z)| = O(1/|z|²). From all this and (*), we have

P ∫ [-∞,∞] 1/(x² + 4)² (x + 1) dx =

= lim [ε→0+] { ∫ [-R,-1-ε] f(x) dx + ∫ [-1+ε,R] f(x) dx } =

= – lim [ε→0+] ∫ [|z+1|=ε] f(z) dz + [R→∞] ∫ [|z|=R] f(z) dz

+ 2πi*Res(z=2i, f(z)) = – { – πi*Res(z=0) } + 0 + 2πi*Res(z=2i, f(z)) =

= 2πi*Res(z=2i, f(z)) + πi*Res(z=0).

(Double minus in the first termafter the equality sign since the integration over the small halfcircle around the pole at z = -1 is in the negative sense.)

( There are some typos, which I hope are rather obvious, e g at the end πi*Res(z=0) should be πi*Res(z=-1,f(z)); the pole is at z = -1. )

Annonser

3 Responses to Principal value integral

  1. No matter if some one searches for his vital thing, therefore he/she wants
    to be available that in detail, therefore that thing is maintained over here.

  2. comedy pictures skriver:

    Haha this page is very amusing. Your website is very
    good. Carry on living it up!

  3. Very soon this website will be famous among all blogging and site-building visitors, ddue
    to it’s fastidious posts

Kommentera

Fyll i dina uppgifter nedan eller klicka på en ikon för att logga in:

WordPress.com Logo

Du kommenterar med ditt WordPress.com-konto. Logga ut /  Ändra )

Google+-foto

Du kommenterar med ditt Google+-konto. Logga ut /  Ändra )

Twitter-bild

Du kommenterar med ditt Twitter-konto. Logga ut /  Ändra )

Facebook-foto

Du kommenterar med ditt Facebook-konto. Logga ut /  Ändra )

Ansluter till %s

%d bloggare gillar detta: