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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Since $\int_1^a \frac{1}{t}\, dt = \int_b^{ab} \frac{1}{u}\, du$ (using the $u$-sub $u = tb$), we have $$\int_1^a \frac{1}{t}\, dt + \int_1^b \frac{dt}{t} = \int_b^{ab} \frac{1}{u}\, du + \int_1^b \frac{1}{t}\, dt = \int_1^b \frac{1}{t}\, dt + \int_b^{ab} \frac{1}{t}\, dt = \int_1^{ab} \frac{1}{t}\, dt.$$

  • $\begingroup$ 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

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:




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.


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 $$ \int_1^a \frac{1}{t}\,dt = \int_b^{ab} \frac{1}{t}\,dt $$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.