1. Week 9 3. Integrals involving functions with branch points
We’ll first calculate some important integrals without branch points.

2. Example 1:
Show that
where t > 0 is a real parameter.
Hint: consider
and change to polar coordinates (r, φ): p = r cos , q = r sin .

3. Example 2:
Show that
where t > 0 and x (of arbitrary sign) are real parameters.
Example 3:
Show that
where t > 0 and x (of arbitrary sign) are real parameters.

4. The inverse Laplace transformation
How do we find L–1[ F(s)] if F(s) isn’t listed in the LT table?...
Theorem 1:
Let F(s) be the Laplace transform of f(t). Then
(1)
where γ is such that the straight line (γ – i∞, γ + i∞) is located to the right of all singular points of F(s).
Comment:
If F(s) → 0 as s → ∞ and t < 0, integral (1) is identically zero due to Jordan’s Lemma (used in yet another alternative formulation).

5. Example 4:
Find
where the branch of s1/2 on the complex s plane is fixed by the condition 11/2 = +1 and a cut along the negative part of the real axis.
Solution:
(2)
where γ > 0.
Close the contour in integral (2) as follows...

6. The integrand is analytic inside the contour – hence,
(3)
Next, let R → ∞, hence
and also (due to Jordan’s Lemma, alt. form.)
Let ε → 0, hence (see TS7, Q4a)

7. Thus, the limiting version of (3) is
hence,
Introduce p > 0:
Recalling how the branch of s1/2 was fixed, we have

8. Then,
hence, interchanging the upper↔lower limits in both integrals,
In the 2nd integral, let p = –pnew, then omit new and ‘merge’ the 1st and 2nd integrals together, which yields

9. Finally, using the result of Example 3, obtain
Comment:
Consider the horizontal segments of the curves C3 and C5 (let’s call these segments C'3 and C'5), and note that Jordan’s Lemma does not guarantee that their contributions vanish as R → ∞.
Note, however, that, for z  C'3,5, the integrand decays as R → ∞, whereas the lengths of C'3 and C'5 remain constant (both equal to γ). Hence, the integrals over C'3 and C'5 do decay as R → ∞ (for rigorous proof, see TS7, Q4b)
Alternatively, we can move γ infinitesimally close to zero: this wouldn’t affect the original integral (why?), but would make the lengths (and, thus, contributions) of C'3 and C'5 equal to zero.