15.7. EVALUATION OF IMPROPER INTEGRALS 387
15.7 Evaluation of Improper IntegralsLetting p(x) ,q(x) be polynomials, you can use the above method of residues to evaluateobnoxious integrals of the form
∫∞
−∞
p(x)q(x)dx ≡ limR→∞
∫ R−R
p(x)q(x)dx provided the degree of
p(x) is two less than the degree of q(x) and the zeros of q(z) involve Im(z) > 0. Theseintegrals are called Cauchy principal value integrals. The contour to use for such problemsis γR which goes from (−R,0) to (R,0) along the real line and then on the semicircle ofradius R from (R,0) to (−R,0).
x
y
Letting CR be the circular part of this contour, for large R,∣∣∣∫CR
p(z)q(z)dz
∣∣∣≤ πR CRk
Rk+2 whichconverges to 0 as R→ ∞. Therefore, it is only a matter of taking large enough R to encloseall the roots of q(z) which are in the upper half plane, finding the residues at these pointsand then computing the contour integral. Then you would let R→ ∞ and the part of thecontour on the semicircle will disappear leaving the Cauchy principal value integral whichis desired. There are other situations which will work just as well. You simply need to havethe case where the integral over the curved part of the contour converges to 0 as R→ ∞.
Here is an easy example.
Example 15.7.1 Find∫
∞
−∞
1x2+1 dx
You know from calculus that the answer is π . Lets use the method of residues tofind this. The function 1
z2+1 has poles at i and −i. We don’t need to consider −i. Itseems clear that the pole at i is of order 1 and so all we have to do is take limz→i
x−i1+x2 =
1(x−i)(x+i) (x− i) = 1
2i . Then the integral equals 2πi( 1
2i
)= π .
That one is easy. Now here is a genuinely obnoxious integral.
Example 15.7.2 Find∫
∞
−∞
11+x4 dx
It will have poles at the roots of 1+ x4. These roots are(12− 1
2i)√
2,−(
12+
12
i)√
2,−(
12− 1
2i)√
2,(
12+
12
i)√
2
Using the above contour, we only need consider −( 1
2 −12 i)√
2,( 1
2 +12 i)√
2. Since theyare all distinct, the poles at these two will be of order 1. To find the residues at these points,you would need
limz→−( 1
2+12 i)√
2
(z−(−( 1
2 −12 i)√
2))
1+ z4 , limz→( 1
2+12 i)√
2
(z−(( 1
2 +12 i)√
2))
1+ z4
As noted above, you could use L’Hospital’s rule to find these limits.
limz→−( 1
2+12 i)√
2
14z3 , lim
z→( 12+
12 i)√
2
14z3