Not always. Take g(x)=x so that we have a Riemann integral. Define f on [0,1] by f(x)=x except f(0)=1. Then 1/f is not bounded on [0,1], thus not Riemann integrable.
2. This is
1/n*\sum_{k=1}^n f(k/n),
where f(x)=x/(1+x^2). So by our problem from HW #2, the limit is \int_0^1 x dx/(1+x^2), which can be evaluated by the FTC as 1/2*log(2).
3(a). We need a function whose derivative exists but is not bounded on [0,1]. Recall that f(x)=x^2*sin(1/x) for x non-zero, f(0)=0, has the derivative 2*x*sin(1/x)-cos(1/x) for x non-zero and 0 at x=0. But this function is bounded, and we need an unbounded function near 0. So if we take instead f(x)=x^2*sin(1/x^2) for x non-zero, f(0)=0, then f'(x)=2*x*sin(1/x^2)-2*cos(1/x^2)/x, f'(0)=0, and f' is not bounded on [0,1].
(b) Let f(x)=0 on [0,1/2), =1 on [1/2,1]. Then f is Riemann integrable on [0,1]. But if f=F' on [0,1], then F' cannot have a singularity of the first kind (see p. 109 of Rudin) at x=1/2, which f has.
4. Clearly f^n is at most M^n so that (\int_0^1 f^n)^(1/n) is at most M.
Next we show that the limit is at least M-e, for any e>0, where e is less than M/2. This implies the limit is M.
Since f is continuous, its maximum value is attained, so choose x_0 in [a,b] such that f(x_0)=M. Snce f is continuous at x_0, there is some interval I about x_0 such that f(x)>M-e on I. (I could have x_0 as an endpoint.) Then \int_0^1 f^n is at least (M-e)^n*length(I), so that taking n th roots we get a lower bound of (M-e)*length(I)^(1/n), which approaches M-e as n-> \infty.
Rudin #13:
(a) The integration by parts, with cos(u) replaced by -1, gives an upper bound of f(x) of
cos(x^2)/2x-cos((x+1)^2)/2(x+1)+1/2*(1/x-1/(x+1))
=1/2x*(1+cos(x^2))+1/2(x+1)*(-1-cos((x+1)^2))
which is at most 1/x.
For a lower bound replace cos(u) by 1 and the lower bound is
cos(x^2)/2x-cos((x+1)^2)/2(x+1)-1/2*(1/x-1/(x+1))
greater than -1/x.
(c) Note that
cos((x+1)^2)-cos(x^2)=2*sin(x^2+x+1/2)*sin(x+1/2)=2*sin(X^2+1/4)*sin(X),
where X=x+1/2. We shall choose X_n=2*n*\pi+\pi/2+e_n, where e_n->0 as n->\infty. Then the second sin factor approaches 1. We show that by suitably choosing e_n, we can make the first factor approach +1 or -1.
(2*n*\pi+\pi/2+e)^2= (2n+1/2)^2*\pi^2+2*e*(2n+1/2)*\pi+e^2 =A+2*e*(2n+1/2)*\pi+e^2.
So if we consider e as a small perturbation, we have available for choosing e_n an interval whose left endpoint is A, of length at least 2*e*(2n+1/2)*\pi for e_n. So if e=1/n, this interval has length at least 4*pi, and we can choose points in this interval with sin=+1 or -1, giving a choice for e_n.
yes. Proof #1: we show that \int_1^R sin(x^2) dx=1/2*\int_1^{R^2} sin(u)/sqrt(u) has a limit as R->\infty by integrating by parts, and then bounding the integral as before,\int_1^\infty du/u^{3/2} does converge.
proof #2: we show that\sum_{n=0}^\infty f(n) converges. f(n)=(cos(n^2)-cos((n+1)^2))/n+O(n^2), so we show that \sum_{n=1}^\infty (cos(n^2)-cos((n+1)^2))/n converges. But this follows from Theorem 3.42, with a_n=cos(n^2)-cos((n+1)^2), b_n=1/n.
Rudin #15:
The equality follows from integration by parts. The Cauchy-Schwarz inequality (see prob #12, with p=q=2) implies that 1/4 is a lower bound. But just like vectors, you have equality in the Cauchy-Schwarz inequality only when the two functions are multiples of one another, this implies that for equality we must have f'(x)=c*x*f(x), or f(x)=K*exp(c*x^2/2), which is never zero at x=a unless K=0, which contradicts f being positive.
Rudin #16: Let's evaluate 1/s*\int_1^N [x]/x^(s+1) dx by integration by parts, dg=d[x], f=1/x^s,
\sum_{k=2}^N 1/k^s=\int_1^N 1/x^s d[x]=\int_1^N s/x^(s+1) [x] dx+(1/N^{s-1}-1)
So
\sum_{k=2}^N 1/k^s=\int_1^N s/x^(s+1) [x] dx+1/N^{s-1},
and we let N->\infty, using s>1.
(b) This is done in the same way.