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Prove that $$\int_{1}^{a} \frac 1t dt + \int_{1}^{b} \frac 1t dt = \int_{1}^{ab} \frac 1t dt$$

Useful facts: $\int_{1}^{a} \frac 1t dt$ can be written as $\int_{b}^{ab} \frac 1t dt$

Every partition $P=${$t_0,...,t_n$} on $[1,a]$ gives rise to a partition $P'=${$bt_0,...,bt_n$} on $[b,ab]$, and conversely.

I have to do the question without actually integrating it, since the Fundamental Theorem of Calculus was not proven yet.

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  • $\begingroup$ Integral of 1/t is log... $\endgroup$ – voldemort Jan 22 '15 at 2:51
  • $\begingroup$ You can just integrate it. $\endgroup$ – Shahar Jan 22 '15 at 2:55
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    $\begingroup$ Your answerers are assuming that this question comes after you have learned the Fundamental Theorem of Calculus, but the $P=$ part of the question indicates you need to compute the integral from scratch using Riemann Sums. Can you clarify? $\endgroup$ – Matthew Leingang Jan 22 '15 at 3:02
  • $\begingroup$ I think I have to use the fact that $$U(f,P)-L(f,P) < \epsilon$$ $\endgroup$ – Senya Jan 22 '15 at 3:03
  • $\begingroup$ Ah, Spivak, my old friend. $\endgroup$ – Matthew Leingang Jan 22 '15 at 3:04
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Let the partitions $P$ and $P'$ be as given.

Notice that

$$ \frac{1}{b} \cdot \text{inf}\{\frac{1}{t} : t_{i-1} \leq t \leq t_i \} = \text{inf}\{\frac{1}{t} : bt_{i-1} \leq x \leq bt_i \}$$

Now we can show equivalence between the two partitions in the following way :

Let the first inf above be $m_i$ and the second be $m'_i$, then we have

$$\begin{align*} L(f, P') &= \sum\limits_{i = 1}^n m'_i (bt_{i} - bt_{i -1}) \\ &= \sum\limits_{i = 1}^n bm'_i (t_{i} - t_{i -1}) \\ &= \sum\limits_{i = 1}^n m_i (t_{i} - t_{i -1}) \\ &= L(f, P) \end{align*} $$

Since these lower step functions are the same, we can conclude that

$$ \text{sup}\{L(f, P)\} = \text{sup}\{L(f, P')\}$$

and thus that $$\int_{1}^{a} \frac{1}{t} dt = \int_{a}^{ab} \frac{1}{t} dt $$

Now we need only plug this in to

$$\int_{1}^{a} \frac{1}{t} dt + \int_{1}^{b} \frac{1}{t} dt = \int_{1}^{ab} \frac{1}{t} $$

and remember that

$$\int_{a}^{b} f + \int_{b}^{c} f = \int_{a}^{c} f $$

to complete the proof.

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  • $\begingroup$ Hi, thank you for your beautiful answer but I wanted to know in the initial step how you can justify the $\frac 1b$ in the third line. I understand what it is intuitively, but does it have to be formally presented or is it just trivial/obvious? $\endgroup$ – Senya Jan 22 '15 at 4:21
  • $\begingroup$ @Senya I had a longer explanation typed up, but the formatting was off. In any case, I think you should be fine taking it as obvious. Perhaps note that the smallest value on an interval in the new partition corresponds the the smallest value on the same $i^{th}$ interval of the old partition, simply multiplied by b. $\endgroup$ – gabe Jan 22 '15 at 4:36
  • $\begingroup$ Thank you ever so much for your help! Also, could a similar proof be used for something similar, such as $\int_a^b f(x) dx = \int_{a+c}^{b+c} f(x-c) dx?$ $\endgroup$ – Senya Jan 22 '15 at 4:37
  • $\begingroup$ @Senya The idea is very similar, yes. You must notice the link between the partitions for the first integral and those of the second. But yes, very similar ideas. $\endgroup$ – gabe Jan 22 '15 at 4:47
  • $\begingroup$ Sorry if I'm asking too much but, what is the link between them? $\endgroup$ – Senya Jan 22 '15 at 4:56
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Since $\int_1^a \frac{1}{t}\, dt = \int_b^{ab} \frac{1}{u}\, du$ (using the $u$-sub $u = tb$), we have $$\int_1^a \frac{1}{t}\, dt + \int_1^b \frac{dt}{t} = \int_b^{ab} \frac{1}{u}\, du + \int_1^b \frac{1}{t}\, dt = \int_1^b \frac{1}{t}\, dt + \int_b^{ab} \frac{1}{t}\, dt = \int_1^{ab} \frac{1}{t}\, dt.$$

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  • $\begingroup$ Sorry, but what is the u-sub? $\endgroup$ – Senya Jan 22 '15 at 2:55
  • $\begingroup$ I mean $u$-substitution. $\endgroup$ – kobe Jan 22 '15 at 2:56
  • $\begingroup$ Ordinarily this would work, but not in the context of this problem which requires a proof from partitions. $\endgroup$ – Matthew Leingang Jan 22 '15 at 3:29
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Abuse your useful fact:

$$\int_{1}^{a} \frac 1t dt + \int_{1}^{b} \frac 1t dt=\int_b^{ab}\frac1t dt+ \int_{1}^{b} \frac 1t dt =-\int_{ab}^{b}\frac1t dt- \int_{b}^{1} \frac 1t dt = -\left(\underbrace{\int_{ab}^{b}\frac1t dt+ \int_{b}^{1} \frac 1t dt}_{\int_{ab}^1\frac1{t}dt}\right)=\;\therefore \int_1^{ab}\frac1{t}dt$$

Other useful facts used:

$$\int_a^bf(t)dt=-\int_b^af(t)dt$$

$$\int_a^bf(t)dt+\int_b^cf(t)dt=\int_a^cf(t)dt$$

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Simply, use the change change of variable $u = b\cdot t$ in the first integral, or $u = a \cdot t$ in the second integral, then use Chasles's relation.

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Following the hint, you want to show that \begin{align} U(f,P) &= U(f,P') \\ L(f,P) &= L(f,P') \end{align} Therefore $$ \int_1^a \frac{1}{t}\,dt = \int_b^{ab} \frac{1}{t}\,dt $$

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  • $\begingroup$ Wait, how is the last part implemented? The fact that $\int_{1}^{a} \frac 1t dt = \int_b^{ab} \frac 1t dt$ $\endgroup$ – Senya Jan 22 '15 at 3:29
  • $\begingroup$ The left hand side is the greatest lower bound of $U(f,P)$, and the right hand side is the greatest lower bound of $U(f,P')$. Does that help? $\endgroup$ – Matthew Leingang Jan 22 '15 at 3:41
  • $\begingroup$ Oh, okay. I think I understand. Thank you for the clarification. $\endgroup$ – Senya Jan 22 '15 at 3:46

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