A Sophism in Calculus

Evaluating indefinite integral \int \frac{1}{x}\;dx using integration by parts gives

\int \frac{1}{x}\;dx =\int x'\cdot \frac{1}{x}\;dx=x\cdot\frac{1}{x}-\int x\cdot(\frac{1}{x})'\;dx=1-\int x\cdot\frac{-1}{x^2}\;dx=1+\int \frac{1}{x}\;dx.

That is,

\int \frac{1}{x}\;dx = 1 + \int \frac{1}{x}\;dx.

Substracting \int \frac{1}{x}\;dx from both sides we conclude


The expression

\int f(x)\;dx = g(x)


\left(\int f(x)\; dx - g(x)\right)' = 0;


\int f(x)\;dx- g(x) = C

where C is a constant.


\int \frac{1}{x}\;dx = 1+ \int \frac{1}{x}\;dx \implies \int \frac{1}{x}\;dx - (1 + \int \frac{1}{x}\;dx) =C.


C = -1.

Moreover, define \leftrightarrow as

f(x) \leftrightarrow g(x), if f(x)-g(x)=C, \quad\quad\quad(2)

it can be shown that

f(x) \leftrightarrow h(x), h(x) \leftrightarrow g(x) \implies f(x) \leftrightarrow g(x) \quad\quad\quad(2-1)


f(x) \leftrightarrow g(x) \Longleftrightarrow f(x)+h(x) \leftrightarrow g(x) + h(x).\quad\quad\quad(2-2)

If we proceed to evaluate \int \frac{1}{x}\;dx as follows:

\int \frac{1}{x}\;dx \leftrightarrow \int x'\cdot \frac{1}{x}\;dx \leftrightarrow x\cdot\frac{1}{x}-\int x\cdot(\frac{1}{x})'\;dx \leftrightarrow 1-\int x\cdot\frac{-1}{x^2}\;dx \leftrightarrow 1+\int \frac{1}{x}\;dx,

we have (by (2-1))

\int \frac{1}{x}\;dx \leftrightarrow 1 + \int \frac{1}{x}\; dx.

Substracting \int \frac{1}{x}\;dx from both sides (by (2-2)) we conclude

0 \leftrightarrow 1.\quad\quad\quad(3)

Unlike (1), (3) is true:

0-1 = -1, a constant.

By the way,

\int \frac{1}{x}\;dx \leftrightarrow \log(x).

See “Introducing Lady L” for details.

Exercise-1 Can you spot the fallacies?

For x>0,

x^2 = x\cdot x = \underbrace{x+x+x...+x}_{x\;\; x's}.

Differentiating both sides, we have

2x=\underbrace{1+1+1+... + 1}_{x\;\;1's}=x.

Dividing both sides by x (since x>0) yields

2 = 1.

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