Problems, set 6. PHYSICS 212A: Mathematical Physics
1. Show that
a)Z2π
0
ecos θcos(nθ −sin θ)dθ =2π
n!b)Z2π
0
ecos θsin(nθ −sin θ)dθ = 0
2. Evaluate for a > b > 0, simplify to elementary function(s) (log, etc.):
I=Z2π
0
ln(a+bcos θ)dθ
Hint. One approach is to use differentiation with respect to a parameter.
3. Evaluate (0 < θ0< π,n= integer):
I=1
πPZπ
0
cos nθ
cos θ−cos θ0
dθ
4. Evaluate the integral (without using variables change x= sin θ) by contour integration. Use
the contour that goes along the upper rim of the negative real axis, around the branch cut,
back to x=−∞ along the lower rim of x-axis for x < 1, and closes by an infinite circle.
I=Z1
−1
dx
√1−x2(x+ 2)
5. Evaluate by contour integration
I=Z∞
0
ln x
x2+ 1 dx .
Hints. Direct the branch cut along the negative x-axis. Design a contour that includes the
upper rim of the branch cut.
6. M&W: 3-18, use the approach and the result of the previous problem.
7. Sum the series
∞
X
n=1
(−1)n+1 1
n2
by converting it into a contour integral using the pole structure of the function π/ sin(πz).
Hint. Extend the sum to −∞.
8. Evaluate, for the contour Cbeing a |z|= 5/2 circle, the integral
I=ZCΓ(z)eaz dz.
9. ∗(Extra credit). Evaluate by contour integration (a) M&W 3-24, (b)
I=Z∞
0
(ln x)2
(x+ 1)2dx .
