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Define $L:(0,\infty)\to\mathbb{R}$ by

$$L(x)=\int_{1}^{x}\frac{dt}{t}.$$

How can I show that $L$ is continuous on $(0,\infty)$?

I know that the definition of continuity states the following:

A function $f:A\to\mathbb{R}$ is continuous at a point $c\in A$ if, for all $\epsilon>0$, there exists a $\delta>0$ such that whenever $|x-c|<\delta$ (and $x\in A$), it follows that $|f(x)-f(c)|<\epsilon$.

If $f$ is continuous at every point in the domain $A$, then we say that $f$ is continuous on $A$.

I have tried applying this definition by letting $x,c\in(0,\infty)$. Then

$$\left|L(x)-L(c)\right|=\left|\int_{1}^{x}\frac{dt}{t}-\int_{1}^{c}\frac{dt}{t}\right|=\left|\int_{1}^{x/c}\frac{dt}{t}\right|=\left|L(x/c)\right|<\epsilon,$$

but no matter how hard I look, I cannot seem to find the connection between that and $|x-c|<\delta$, because these are integrals, and I cannot perform algebraic manipulations to reach the desired inequality.

Do you guys have any ideas?

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2  
We want to show that by suitable choice of $\delta$, we can make the absolute value of $\int_1^{x+\delta}\frac{dt}{t} -\int_1^x\frac{dt}{t}$ small. So we are interested in the absolute value of $\int_x^{x+\delta}\frac{dt}{t}$. This is less than $\delta$ times the max of $\frac{1}{t}$ in our interval. –  André Nicolas May 1 '12 at 4:29
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Have you covered the fundamental theorem of calculus in your class yet? If so, you could just show that the derivative of L exists on your interval and that will imply continuity. –  Nishrito May 1 '12 at 4:41
    
In fact, your function is the logarithmic function ; you should not only expect continuity, but even infinite differentiability... –  Patrick Da Silva May 1 '12 at 4:46
    
@Pat: you and I know that, but it looks to me that OP is working from this definition and pretending that he knows nothing about the logarithm. –  J. M. May 1 '12 at 5:02
    
@J.M. : It's not because it's a different definition that there's no way to prove this, even if you know nothing about the logarithm! It's not because you pretend you know nothing that you actually do know nothing... hehe =P –  Patrick Da Silva May 1 '12 at 7:22

4 Answers 4

up vote 3 down vote accepted

The proofs one can use depend on what has been proved so far. There are two alternatives. One could question you about whether various theorems have been proved, in your course, or one could guess. I will guess that only very basic properties of the definite integral have been proved rigorously.

Let $c\in (0,\infty)$. We show that $L$ is continuous at $c$. That means that for any $\epsilon \gt 0$, we must produce (or at least show the existence of) a $\delta \gt 0$ such that if $|x-c| \lt \delta$, then $|L(x)-L(c)| \lt \epsilon$. Note that $$|L(x)-L(c)|=\left|\int_1^x\frac{dt}{t} -\int_1^c\frac{dt}{t}\right|=\left|\int_c^x\frac{dt}{t}\right|.$$ We estimate the last integral. If $\frac{1}{t} \lt M$ in the interval from $c$ to $x$ (but $x$ could be to the left of $c$), we have $$\left|\int_c^x\frac{dt}{t}\right| \le |x-c|M.$$ We first of all make $\delta<c/2$. That makes sure that $t$ does not dip below $c-c/2=c/2$, so in the interval of integration, $\frac{1}{t}$ is bounded above by $M=\frac{2}{c}$.

If we make that restriction on $\delta$, then if $|x-c| \le \delta$, we have $$\left|\int_c^x\frac{dt}{t}\right| \le |x-c|M \le \delta\frac{2}{c}\tag{$\ast$}.$$

Make $\frac{2\delta}{c} \lt \epsilon$, that will do it! So let's make $\delta \lt c\epsilon/2$. For definiteness, make $\delta=c\epsilon/5$, why not? That's not quite right, remember that we wanted $\delta<c/2$. So make $\delta$ equal to the minimum of $c/2$ and $c\epsilon/5$. This ties down everything. Part of it was overkill, as protection against the possibility that as a joke we are given $\epsilon=1000$. If we look only at small $\epsilon$, say less than $1$, then $\delta=c\epsilon/5$ is plenty good enough, indeed $\delta=c\epsilon/2$ is good enough.

The paragraph above is of very slight importance. In essence, once we had the inequality $(\ast)$, it was clear that by appropriate choice of $\delta$ we could make $|L(x)-L(c)|$ as small as we wished.

Remark: The approach you tried also works, and the details are not all that different. It requires first proving the fact that the difference is $\int_1^{x/c}\frac{dt}{t}$, but I imagine that was done in your course. (That special fact would not be available if we had say $1/\sqrt{t}$ instead of $1/t$.) Then we can find an upper bound for the absolute value, of the shape $|x/c-1|M$, where $M$ is as in the solution above. Finally, we need to estimate $|x/c-1$ in terms of $|x-c|$, that part is easy.

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$f$ be integrable on the $[a,b]$, then $F=\int_{a}^{x}f(t)dt$ is continuous apply this result to prove L is continuous

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To show continuity on $(0,\infty)$, it is sufficient to show that $L$ is continuous on the interval $(a,\infty)$ for any $a>0$. The advantage of doing this is that you can easily bound $t \mapsto \frac{1}{t}$ on the interval $(a, \infty)$ by $\frac{1}{a}$.

The we have, for $x,y \in (a, \infty)$, $$| L(x)-L(y) | = |\int_y^x \frac{1}{t} dt | \leq \frac{1}{a} | x-y|.$$ Choosing $\delta < a \epsilon$, then you have the required result, ie, if $|x-y| < \delta$, then $| L(x)-L(y) | \leq \frac{1}{a} \delta < \epsilon$.

Since $a>0$ was arbitrary, you can conclude that $L$ is continuous on $(0,\infty)$.

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You did something stronger than prove continuity ; you proved uniform continuity of $L$ over any subset of $(0,\infty)$ that does not have $0$ as a limit point (because that is where the problem is with the logarithm function). You didn't need to be this much violent to prove continuity. (Although I like the use of sledgehammers in proofs, they're fun to look at. =P) +1 –  Patrick Da Silva May 1 '12 at 7:22

Take a close interval containing c, there the function f(t)=1/t is bounded. Then use monotony of integral between c to x. (For x sufficient close to c you are working in the close interval you took).

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This is to finilize your prove, but, as other people posted you can use the fundamental theorem. –  alpha.Debi May 1 '12 at 5:10
    
"finilize" is not an english word.... perhaps "complete" or "finish" are more appropriate? –  Patrick Da Silva May 1 '12 at 7:18

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