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I'm having trouble w/ the following result, which is Ch 6, Thm 15, in Buck's Advanced Calculus:

If $f(x,t)$ is continuous on $x\geq b$,$t\geq c$, $\int_c^\infty f(x,t)\;dt=F(x)$, uniformly on $x\geq b$, and $\lim_{x\to \infty} f(x,t)=g(t)$ uniformly on $t\in[c,L]$, $\forall L\geq c$. Then $\lim_{x\to \infty}F(x)=\int_c^\infty g(t)\;dt$.

In other words, we can swap the limit and the integral.

Buck's Proof: Fix $\epsilon>0$. Since $\int_c^\infty f(x,t)\;dt=F(x)$, uniformly on $x \geq b$, choose $r\geq c$ s.t. $\left|F(x)-\int_c^rf(x,t)\;dt\right|< \epsilon$.

Since $\lim_{x\to \infty}f(x,t)=g(t)$ uniformly on $t\in[c,r]$, choose $x_0\geq b$ s.t. $|f(x,t)-g(t)|<\frac{\epsilon}{r-c}$ $\forall x\geq x_0,t\in[c,r]$.

Then if $x_1, x_2 > x_0$, $|F(x_1)-F(x_2)|<4\epsilon$, by a Triangle Inequality argument.

Thus, F(x) has the Cauchy property as $x\to \infty$, so $\lim_{x \to \infty}F(x)$ exists, and the desired result follows.

I'm really confused why the desired result, $\lim_{x\to \infty}$F(x)=$\int_c^\infty g(t)\;dt$, follows just from showing the LHS exists.

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Once you establish that $\lim_{x \to \infty} F(x) = L$ exists, then you can show that $g$ is integrable on $[c,\infty)$ and

$$\int_c^{\infty} g(t) \, dt = L.$$

Note that

$$\left|\int_c^r g(t) \, dt - L\right| \leqslant \left|\int_c^r g(t) \, dt - \int_c^r f(x,t) \, dt\right| + \left|\int_c^r f(x,t) \,dt - \int_c^{\infty} f(x,t) \, dt\right| + \left|\int_c^{\infty} f(x,t) \,dt - L\right|.$$

By uniform convergence of $f(x,t)$ to $g(t)$ and the convergence of $F(x)$ to $L$, the first term and the third term on the RHS are each smaller than $\epsilon/3$ if $x$ is sufficiently large. The second term on the RHS is also smaller than $\epsilon/3$ if $r$ is sufficiently large for any $x \in [b,\infty)$.

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Let $A = \lim_{x\to\infty}F(x)$.

$$ \forall \epsilon > 0 , \, \, \exists r_0 > c, \quad\mbox{s.t.}\quad \forall r > r_0, |F(x) - \int_c^r f(x,t)\mathrm{d}t| < \epsilon$$.

Fix an $r > r_0$, then

$$\forall \epsilon > 0 , \, \, \exists N > 0\quad\mbox{s.t.}\quad \forall x > N, |F(x) - A| < \epsilon$$.

$$ \forall \epsilon > 0 , \, \, \exists M(r) > 0\quad\mbox{s.t.}\quad\forall x > M(r), |f(x,t) - g(t)| < \frac{\epsilon}{r - c}$$.

So $ \forall x > \max(M(r),N)$, $$\begin{align} |\int_c^r g(t)\mathrm{d}t - A| &\le |\int_c^r g(t)\mathrm{d}t - \int_c^r f(x,t)\mathrm{d}t| + |\int_c^r f(x,t)\mathrm{d}t - F(x)| + |F(x) - A| \\ &\le (r - c) \cdot \frac{\epsilon}{r-c} + \epsilon + \epsilon\\ &= 3\epsilon \end{align}$$.

Since for all $\epsilon > 0$, we can find $r_0$ s.t. for all $r > r_0$ we have the means to bound $|\int_c^r g(t)\mathrm{d}t - A|$ within $3\epsilon$, we can conclude that $\int_c^\infty g(t)\mathrm{d}t = A$.

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