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In a question I've got, I found that I need to prove the following equality:

$\lim_{h\rightarrow 0}\int_{0}^{\infty}\frac{e^{-(t+h)x}-e^{-tx}}{hx}dx=\int_{0}^{\infty}\lim_{h\rightarrow 0}\frac{e^{-(t+h)x}-e^{-tx}}{hx}dx$ ($t>0$ is a constant.)

I know the limit in the right side is $-e^{-tx}$ if that matters.

Now, by Heine, it's sufficient to prove this result with a sequence $h_n\rightarrow 0$ instead of a continuous variable. So let's define $f_n(x)=\frac{e^{-(t+h_n)x}-e^{-tx}}{h_nx}$ and now we need to show that the conditions of DCT hold, and all that's left is finding a dominating function.

That's where I failed. Any idea?

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    $\begingroup$ Since $\lim_{h_n\to 0}\big|\frac{e^{-h_nx}-1}{h_nx}\big|=1$, then for sufficiently large $n$ we have $\big|\frac{e^{-h_nx}-1}{h_nx}\big|<2$. You could take the dominating function to be $2e^{-tx}$. $\endgroup$ – M. T Aug 30 '16 at 15:55
  • $\begingroup$ Thank you! didn't see that $\endgroup$ – 35T41 Aug 30 '16 at 16:00
  • $\begingroup$ @35T41 My answer is not complete . . . see the one by Dr. M V for a complete one $\endgroup$ – user288742 Aug 31 '16 at 1:21
  • $\begingroup$ I would prefer if you unaccepted mine and accepted his $\endgroup$ – user288742 Aug 31 '16 at 1:29
  • $\begingroup$ Done. I actually used M.T's answer eventually... but thank you anyway $\endgroup$ – 35T41 Aug 31 '16 at 8:01
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A very efficient way forward circumvents use of the Dominated Convergence Theorem and instead exploits Frullani's Theorem. Proceeding, we obtain

$$\frac1h \int_0^\infty \frac{e^{-(t+h)x}-e^{-tx}}{hx} \,dx=\frac1h \log\left( \frac{t}{t+h}\right)$$

The limit as $h\to 0$ is straightforward to evaluate with the answer $-\frac1t$. And we are done!

If one wishes to proceed through use of the Dominated Convergence Theorem, then the following primer will facilitate analysis.

PRIMER:

In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities

$$\bbox[5px,border:2px solid #C0A000]{1+x\le e^x\le \frac{1}{1-x}} \tag 1$$

for $x<1$.


Herein, we use the inequality on the left-hand side of $(1)$ to establish a coveted dominating function. To that end, we proceed.

Note that we can write the integrand $\frac{e^{-(t+h)x}-e^{-tx}}{hx}$ as

$$\begin{align} \frac{e^{-(t+h)x}-e^{-tx}}{hx}&=e^{-tx}\left(\frac{e^{-hx}-1}{hx}\right) \tag{2a}\\\\ &=e^{-(t+h)x}\left( \frac{1-e^{hx}}{hx} \right) \tag{2b} \end{align}$$

Note that the term in parentheses on the right-hand side of $(2a)$ is non-positive for all $x\in [0,\infty$ and all $h$. Applying the left-hand side inequality in $(1)$ to $(2a)$, we find immediately that for $h>0$

$$\left|\frac{e^{-hx}-1}{hx}\right|\le 1$$

for $x\ge 0$, $h>0$.

For $h<0$, we take $|h|<t/2$. Then, applying the left-hand side inequality in $(1)$ to $(2b)$, we see analogously that

$$\left| \frac{1-e^{hx}}{hx} \right|\le 1$$

for $x\ge 0$, $h<0$.

Hence, we have

$$\left|\frac{e^{-(t+h)x}-e^{-tx}}{hx}\right|\le e^{-tx/2}$$

for all $x\ge 0$ and all $|h|\le t/2$.

Inasmuch as $\int_0^\infty e^{-tx/2}\,dx=\frac2t <\infty$, the Dominated Convergence Theorem guarantees that

$$\begin{align} \lim_{h\to 0}\int_0^\infty \frac{e^{-(t+h)x}-e^{-tx}}{hx}\,dx&=\int_0^\infty \lim_{h\to 0}\left(\frac{e^{-(t+h)x}-e^{-tx}}{hx}\right)\,dx\\\\ &=-\int_0^\infty e^{-tx}\,dx\\\\ &=-\frac1t \end{align}$$

And we are done!

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  • $\begingroup$ I have been thinking about this question . . . Is there an issue with the dominating function only holding for $|h| \le t/2$? I guess I am not familiar with the intricacies of LDCT when you don't have a limit as $n \to \infty$ of an integral of some sequence $(f_n)_{n=1}^\infty$. $\endgroup$ – user288742 Aug 31 '16 at 3:08
  • $\begingroup$ Good question. There is no conceptual problem ensuring that $|h|<t/2$ since we eventually let $h \to 0$. $\endgroup$ – Mark Viola Aug 31 '16 at 3:11
  • $\begingroup$ Ok. I thought issues might arise, since I know that in the "sequential version" you must have $|f_n(x)| \le g(x)$ for all $n$ in order to say $g$ is a dominating function. Or maybe this only needs to hold for sufficiently large $n$ as well? $\endgroup$ – user288742 Aug 31 '16 at 3:15
  • $\begingroup$ The latter statement is correct. We only require that there is an $N$ such that $|fn(x)|\le g(x)$ for all $n>N$. $\endgroup$ – Mark Viola Aug 31 '16 at 3:22
  • $\begingroup$ Thanks - that is good to know ! $\endgroup$ – user288742 Aug 31 '16 at 3:24
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Fix $h >0$. Note that $$ \frac{e^{-(t+h)x}-e^{-tx}}{hx} = e^{-tx} \left( \frac{e^{-hx}-1}{hx} \right) \implies \left\vert \frac{e^{-(t+h)x}-e^{-tx}}{hx} \right\vert = e^{-tx} \left\vert \frac{e^{-hx}-1}{hx} \right\vert $$

Now, define $f(x) = e^{-x}$. Then $$\frac{e^{-hx}-1}{hx} = \frac{f(hx) - f(0)}{hx}$$ and by Mean Value Theorem $$ \frac{f(hx) - f(0)}{hx} = f'(\xi_x) = -e^{-x} \Big|_{x = \xi_x} = -e^{-\xi_x} $$ for some $\xi_x \in (0,hx)$. Now define $g(y) = -e^{-y}$ for $y>0$. It is clear that $|g(y)| < 1$ for all $y >0$.

So for all $x$ we have $$ \left\vert \frac{ f(hx) - f(0)}{hx} \right\vert = |-e^{-\xi_x}| \le 1 $$

which means $$ \left\vert \frac{e^{-(t+h)x}-e^{-tx}}{hx} \right\vert = e^{-tx} \left\vert \frac{e^{-hx}-1}{hx} \right\vert \le e^{-tx} $$ and the RHS is integrable on $[0,\infty)$.

EDIT: If $h<0$, then write $$ \frac{e^{-(t+h)x}-e^{-tx}}{hx} = e^{-(t+h)x} \left( \dfrac{ 1 - e^{hx}}{hx} \right)$$ and $$ \dfrac{ 1 - e^{hx}}{-hx} = \dfrac{ g(0)- g(hx)}{-hx} $$ where $g(x) = e^{x}$. Then, by MVT we get $$\dfrac{ g(0)- g(hx)}{-hx} = g'(\xi_x) = e^{\xi_x} $$ for some $\xi_x$ in $(hx,0)$. Then, $e^{\xi_x} \le 1$ because of this interval.

We can further restrict ourselves to $-\frac{t}{2} < h < 0$, since we're taking a limit and only care about small $h$. Finally $$ \left\vert e^{-(t+h)x} \left( \dfrac{ 1 - e^{hx}}{hx} \right) \right\vert \le e^{-(t+h)x} \le e^{-tx/2} $$ and this serves as a dominating function.

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    $\begingroup$ This doesn't apply when $h<0$. So, you have proven convergence from the right only. ;-) $\endgroup$ – Mark Viola Aug 30 '16 at 21:21
  • $\begingroup$ I would delete but I can't since it's accepted. My apologies $\endgroup$ – user288742 Aug 31 '16 at 1:28

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