A fun time — in mathematics.
This one sophism is using integral and it is taken from Mudd Math Fun Facts — the calculation ends up with 0 = 1. The link is also gone now, as they revamped the website.
Anyway, sophism means a fallacious argument, especially one used deliberately to deceive.
Sophisticate originates from sophisticus (Mediaeval Latin), while sophism is from Greek sophisma — different origins, but lo! Sophism and sophisticate, both have sophis. And yes, sophism has its sophistication.
Quoted
Let's continue the last line:
$∫{1/x} dx = 1 + ∫{1/x} dx$
Subtract both sides by $∫{1/x} dx$
$∫{1/x} dx$ $- ∫{1/x} dx$$ = 1 + ∫{1/x} dx$ $- ∫{1/x} dx$
$0 = 1$
❓❓
Those are NOT supposed to be that way.
If we have $∫{1/x} dx$ to be solved using integration by parts, then:
$∫udv = uv - ∫vdu$
$∫udv = ∫{1/x} dx$
Thus:
$u = 1$
$dv = {1/x} dx$
$du = 0$
$v = ∫{1/x} dx$
Plug those in:
$∫{1/x} dx = u.v - ∫v.du$
$∫{1/x} dx = 1.∫{1/x} dx - ∫ (∫{1/x} dx) . 0$
$∫{1/x} dx = 1.∫{1/x} dx - 0$
$∫{1/x} dx = ∫{1/x} dx$
Therefore:
$∫{1/x} dx = ∫{1/x} dx$
Subtract both sides by $∫{1/x} dx$:
$∫{1/x} dx$ $- ∫{1/x} dx$$ = ∫{1/x} dx$ $- ∫{1/x} dx$
$0 = 0$
0 = 0Mate, I've just explained that.
But anyway, if we insist on doing that, it will again produce 0 = 1.
Therefore, integration by parts technique can't be applied to solve that. Well, it can be used, but... it will LOOP BACK to itself.
Properly Referenced
By looking at integral table, we will see this reference:
$∫ 1/x\ dx = ln\ |x| + C$
It is defined as that. In the table. ✨
Well, it's not literally a definition in the formal mathematical sense, it's more the final established result of a derivation that's been promoted to "reference entry" in an integral table.
In teaching and lookup contexts, it's treated like a definition because we just take it from the table without re-deriving it every time.
It's a theorem rather than a definition.
Thus, we need to memorise that table. Mm.
We can always derive it, but then, we will also use a DEFINITION to get to that form.
Let me try to derive it.
It starts from COMPOUND INTEREST. Indeed!
Compound interest is this:
$(1+1/n)^n$
When we put extremely LARGE value for $n$:
$(1+1/n)^n$ as $n→∞$
It will yield ${2.71828...}$ pattern.
Then there was Leonhard Euler to cement the $e$. It's the ${Euler's}$ ${cheek}$.
$e$ is ${2.71828...}$ — the mysterious number. It's under ${irrational}$ ${numbers}$.
${IRRATIONAL}$
${foo}$ ${foo}$ ${twaddle}$ ${twaddle}$ ${narf}$ $∴$ $e$$∴$ means "therefore".
$ln$ is $log_e$ ➡️ "${logarithm}$ to base $e$".
It is a Mediaeval Latin term ➡️ $l$ and $n$ ➡️ $ln$ ⬇️
$l$ogarithmus $n$aturalis.
Indeed, not nearly Mandarin.
It's "natural" because it is the ${Euler's}$ ${cheek}$. Posterior, it naturally happens on a human (or other mammals), comes in pair.
Jest aside, back in 1600s — it started from the bankers', not bonkers', "curiosity". One sod presumably said:
Hang on, what if we compound more often? Monthly, daily, hourly... infinitely often!
Can we charge people for interest that grows continuously?
So they let $n → ∞$. UNLEASH the... horizontal 8!
Well.
Surprise!
Instead of exploding into chaos, that expression stabilises at a neat limit:
${lim}↙{n → ∞} (1+1/n)^n = e$
Or, let's just flip that:
$e = {lim}↙{n → ∞} (1+1/n)^n$
Resuming the derivation. Let's bridge that limit form above to the integral form. This integral form:
$∫ 1/x\ dx = ln\ |x| + C$
Deriving:
$e = {lim}↙{n → ∞} (1+1/n)^n$ [➡️ BRIDGE ➡️] $∫ 1/x\ dx = ln\ |x| + C$
Right. $e = {lim}↙{n → ∞} (1+1/n)^n$ is not a function, it's a constant. We need the generic function.
FIRST.
From the limit definition of $e$ above, the function $f(x) = e^x$ can be expressed as:
$e^x = {lim}↙{n → ∞} (1+{x/n})^n$
As a side note, that form, if we plug $x = 1$, it will be $e = {lim}↙{n → ∞} (1+1/n)^n$. That form above is essentially compound interest at rate $ðĨ$. The generic function we need. Moving on.
Second. Differentiating $e^x$ directly yields the unique property:
${d/{dx}}(e^x) = e^x$
⬆️ That's a THEOREM.
ð♂️ WHY? Why is the derivative of $e^x$ with respect to $x$ $e^x$?
${lim}↙{h → 0} {b^h - 1}/h = 1$
When $b = e$, because $ln(e) = 1$, the limit equals $1$.
It's the characteristic base property.
Thus:
${lim}↙{h → 0} {b^h - 1}/h = ln(b)$
Let's step back a bit.
For any function $f(x)$, calculus defines its derivative — the slope at any point $x$ — as:
$f'(x) = {lim}↙{h → 0}{f(x + h) - f(x)}/h$
Plugging in $f(x) =$ $e^x$ and $f(x + h) =$ $e^{x + h}$ for the right side:
$f'(x) =$ ${lim}↙{h → 0} {e^{x + h} - e^x}/h$
Substitute $f'(x)$ with ${d/{dx}}(e^x)$ for the left side:
${d/{dx}}(e^x)$ $= {lim}↙{h → 0} {e^{x + h} - e^x}/h$
${d/{dx}}(e^x) = {lim}↙{h → 0} {e^x(e^h - 1)}/h$
${d/{dx}}(e^x) = e^x ⋅ $ ${lim}↙{h → 0} {e^h - 1}/h$
By the DEFINITION of $e$, ${lim}↙{h → 0} {e^h - 1}/h = 1$, therefore:
${d/{dx}}(e^x) = e^x ⋅$ $1$
${d/{dx}}(e^x) = e^x$ ðš
Such a famous superstar in calculus.
Resuming the bridge construction here.
That second step, ${d/{dx}}(e^x) = e^x$ is our TOOL.
Third. Set $y = \log_e(x)$ (assuming $x > 0$). Rewrite it in exponential form.
From this form:
— to this exponential form:
Fourth. Differentiate both sides with respect to $x$:
${d/{dx}}$$(e^y) = $${d/{dx}}$$(x)$
The tool ${d/{dx}}(e^x) = e^x$ applies at the differentiation of $e^y$ with respect to $x$:
$e^y ⋅ {{dy}/{dx}}$$ = 1$
âđ️ ${d/{dx}}(e^y) = $$e^y ⋅ {{dy}/{dx}}$
Fifth. Substitute $e^y = x$ back into the equation:
$e^y ⋅ {{dy}/{dx}} = 1$
$x$$ ⋅ {{dy}/{dx}} = 1 ⟹ {{dy}/{dx}} = {1/x}$
Sixth. Since $y = \log_e(x)$, thus:
${{dy}/{dx}} = {1/x}$
${d/{dx}}($$\log_e(x)$$) = {1/x}$
Flip that:
${1/x} = {d/{dx}}(\log_e(x))$
Seventh. Integrate both sides:
$∫$ ${1/x}$ $dx$ =$∫$ ${d/{dx}}(\log_e(x))$$dx$
Eighth. Because integration and differentiation are inverse operations (fundamental theorem of calculus), the integral cancels out the derivative on the right side. This part:
$∫ {d/{dx}}(\log_e(x))dx$
— becomes:
$log_e(x) + C$
Therefore:
$∫{1/x} {dx} = $$log_e(x) + C$
Ninth. Since ${1/x}$ is defined for negative numbers but $log_e(x)$ is ONLY defined for $x > 0$, the domain is extended by placing absolute value bars $|x|$:
$∫{1/x} {dx} = log_e\ $$|x|$$ + C$
Or:
Done. ✅
Right.
Bridge emoji, activate!
$e = {lim}↙{n → ∞} (1+1/n)^n$
ð
$e^x = {lim}↙{n → ∞} (1+{x/n})^n$
${d/{dx}}(e^x) = e^x$
$y = \log_e(x)$
$e^y = x$
$...$
${and}$ ${such}$
ð
$∫{1/x} {dx} = ln\ |x| + C$
Hm.
"Compound interest" but with huge n ➡️ "compound interest".
Ah. Bankers.
By the way, back to our theme — the 0 = 1 steps above are a reference fallacy.
A self-referential definition disguised as a calculation.
Which reference bit?
Er. Again. Mate, I've just shown that. Above, with the bridge emojis. The chain of bridge emojis.
But if you look closely.
Oh, well.
So. Let's make an analogy. In form of a skit. In. A classroom. Possibly.
As such:
(Teacher) That thing over there is called a chair.
(Pupil) Why?
(Teacher) Well, we name it that.
(Pupil) Why?
(Teacher) Because "hair" is taken.
(Pupil) Why?
(Teacher) ðĪĶ Because...ð♂️➡️
(Teacher) (From afar, shouting.) Hahaha, this is called far! (Waving.)
(Pupil) ð
Right. As shown above.
In calculus sense, this form:
$∫ 1/x\ dx = ln\ |x| + C$
— is because the derivative of —
$ln|x|$
— with respect to $x$ is —
$1/x$
❗
ðĪ Well.
Once more:
If:
${d/{dx}}F(x) = f(x)$
Then:
$∫f(x) dx = F(x) + C$
In one frame:
${d/{dx}}F(x) = f(x)$ $⇔$ $∫f(x) dx = F(x) + C$
$⇔$ means "is logically equivalent to".
A because B, and B because A.
AGAIN. Once more.
Since:
$∫ 1/x\ dx = ln\ |x| + C$
So — or, because:
${d/{dx}}ln\ |x| = 1/x$, for $x ≠ 0$
It's just swirling around. It's like:
That colour is red.
So or because:
Red is the colour of that.
Or this tautology:
The parent of Bob is Karen so or because Karen is Bob's parent.
Er. Well.
When we go deeper:
$ln(x)$ := $∫_1^x {1/t} dt$
:= means "is defined as".Mm. The definition.
The Proof
Let's refine that parent-of-Bob tautology.
Integral chair is not banana, because the derivative of banana is not chair. ðĪ·
$∫$ chair $dx = F(x) + C ≠ ∫$ banana $dx$
Therefore, $0 ≠ 1$.
ðĪ