You may use Gamma or Beta function to express the final answer. Please show intermediate steps, otherwise there are no points.

1. Compute the integral (5 points) $\int_0^1 \frac{x^4}{\sqrt{1-x^3}} dx$

Replace variable $u = x^3$, then $I = \int_0^1 \frac{u^{4/3}}{(1-u)^{1/2}} du^{1/3} = (1/3) \int_0^1 (1-u)^{-1/2} u^{2/3}du = (1/3) \int_0^1 (1-u)^{-1/2} u^{2/3}du = (1/3)B(5/3, 1/2).$

2. Compute the integral (5 points) $\int_0^1 x^2 (-\ln x)^2 dx$ Replace variable $u = -\ln x$, then $\int_0^\infty e^{-2u} u^2 e^{-u} du = (1/3)^3 \Gamma(3).$

3. Compute the special value of Gamma functions (the result should not be expressed using Gamma function).

• $\Gamma(-1/2) = \Gamma(1/2)/(1/2) = 2 \sqrt{\pi}$ (2 points)
• $|\Gamma(1/2 + i)| = ?$ (3 points) [Hint: $\Gamma(\bar z) = \overline{\Gamma(z)}$, and $\Gamma(p)\Gamma(1-p) = \pi / \sin(p \pi)$ ]

Take $p=1/2+i$, then $|\Gamma(1/2 + i )|^2 = \Gamma(1/2+i) \Gamma(1/2-i) = \pi / \sin(\pi (1/2 + i)) = \pi / \cos(i \pi) = \pi / \cosh(pi)$

1. Compute $P_3(x)$ using Rodrigue formula. (5 points)

By Rodrigue formula $P_3(x) = \frac{1}{2^3 3!} \d_x^3 (x^2-1)^3 = \frac{1}{2^3 3!} \d_x^3 (x^2-1)^3 = \frac{1}{2^3 3!} \d_x^3 (x^6- 3 x^4 + 3 x^2 - 1) = \frac{1}{48}(120 x^3 - 72 x) = (5 x^3 - 3 x)/2.$

2. Prove the recursion relation 5.8( c) (10 points) $P_l'(x) - xP_{l-1}'(x) = l P_{l-1}(x)$ using the generating function $\Phi(x,h) = \sum_{n=0}^\infty h^n P_n(x) = \frac{1}{\sqrt{1 - 2 x h + h^2} }$

If we multiply both sides of the equation by $h^l$ and sum over $l$, we get $LHS = \d_x \Phi(x,h) - x h \d_x \Phi(x,h)$ $RHS = h \d_h (h \Phi(x,h))$ We plug in $\Phi(x,h)$ to test whether this is true. $LHS = (1-xh) \frac{h}{(1 - 2 x h + h^2)^{3/2}}$ and $RHS = h \Phi + h^2 (-1/2) (-2x+2h) (1 - 2 x h + h^2)^{-3/2} = [h(1-2xh + h^2)+h^2(x-h)](1 - 2 x h + h^2)^{-3/2}$ $= h(1-hx)(1 - 2 x h + h^2)^{-3/2}$ OK, the same, done.

3. Compute $\int_{-1}^1 x^n P_n(x) dx$ in the following steps: ( 10 points)

• Find the constant $c$, such that $P_n(x) = c x^n + \z{ lower order terms}$. (try Rodrigue formula to get the leading term)
• Show that $\int_{-1}^1 P_n(x)^2 dx = \int_{-1}^1 c x^n P_n(x) dx$
• Look up $\int_{-1}^1 P_n(x) P_n(x) dx$.

The leading term of $P_n(x)$ can be obtained as $\frac{1}{2^n n!} (2n)(2n-1)\cdots (n+1) = 2^{-n} {2n \choose n}$

Because lower order term is orthogonal to $P_n(x)$.

$\int_{-1}^1 P_n(x) P_n(x) dx = \frac{2}{2n+1}$

So finally, we get $\frac{2^{n+1}(n!)^2}{(2n+1) (2n)!}$

1. Problem 12.1. Show by ratio test that the series for $J_p(x)$ converges for all $x$. (10 points)

2. Problem 15.6 (7 points)

3. Problem 19.1 (7 points)

4. Problem 20.3, 6, 7 (6 points)

1. Solve the steady state heat equation on 2D square $[0,1]^2$. (10 point) $\Delta u(x,y) = 0, \quad 0 \leq x, y \leq 1$ with boundary condition that $u(0, y)= 0, u(1,y)=1, u(x, 0)=0, u(x,1)=1.$

Solution $u(x,y) = \sum_{n \z{ odd} } \frac{4}{n \pi \sinh(n \pi)} \left( \sin(n \pi x) \sinh(n \pi y) + \sin(n \pi y) \sinh(n \pi x) \right)$

2. Solve the steady state heat equation on 3D unit ball. (10 point) $\Delta u(r, \theta, \phi) = 0$ with boundary condition at $r=1$ that $u(1, \theta, \phi) = \cos(\theta) \sin(\theta) \sin(\phi)$

Hint: use $P_2^1(\cos \theta) = -3 \cos(\theta) \sin(\theta)$

Solution: $u = (-1/3) r^2 P_2^1(\cos(\theta)) \sin(\phi)$

3. Solve the heat flow equation on a circle. (10 point) $\d_t u(t, \theta) = \d_\theta^2 u(t, \theta).$ such that the initial condition is $u(0, \theta) = \cos^2(\theta).$

$u(\theta, t) = e^{-4t} (1/2) \cos(2\theta) + 1/2$