# 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 $0=1.\quad\quad\quad(1)$

The expression $\int f(x)\;dx = g(x)$

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

i.e., $\int f(x)\;dx- g(x) = C$

where $C$ is a constant.

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

Or, $C = -1.$

Moreover, define $\leftrightarrow$ as $f(x) \leftrightarrow g(x),$ if $f(x)-g(x)=C, \quad\quad\quad(2)$

it can be shown that $f(x) \leftrightarrow h(x), h(x) \leftrightarrow g(x) \implies f(x) \leftrightarrow g(x) \quad\quad\quad(2-1)$

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

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 $2 = 1.$