Integration in a non-standard way: evaluating $ \int \frac 1 { x ^ 2 } \, \mathrm d x $ without applying the power rule - Is it nonsense?

Question

I want to evaluate the integral $$ \int \frac 1 { x ^ 2 } \, \mathrm d x $$ in a non-standard way and without applying the power rule.

Let $ x \ne 0 $ and $$ \int \frac 1 { x ^ 2 } \, \mathrm d x = f ( x ) + C \text . $$ For any $ a \in \mathbb R \setminus \{ 0 \} $ we have $$ f ( a x ) = \int \frac 1 { a ^ 2 x ^ 2 } \, \mathrm d ( a x ) = \frac 1 a \int \frac 1 { x ^ 2 } \, \mathrm d x = \frac 1 a \bigl ( f ( x ) + C \bigr ) \text . $$

Then, we have $$ f ( x ) = a f ( a x ) \text , $$ for any $ a \in \mathbb R \setminus \{ 0 \} $.

Putting $ x = 1 $, we have $$ f ( a ) = \frac { f ( 1 ) } a \text . $$

This is the critical point for me. I'm not sure I'm not wrong here. I will just continue.

Here, I will replace the constant $ a $ with the variable $ x $. My argument for doing this is as follows: $ a $ covers all nonzero real numbers.

So, I will write the last equation like this (but, I'm not sure): $$ f ( x ) = \frac { f ( 1 ) } x \text . $$

We continue.

Let $ u = x $ and $ v = f ( x ) $, so that $ \mathrm d u = \mathrm d x $ and $ \mathrm d v = \frac 1 { x ^ 2 } \, \mathrm d x $.

$$ \int x \cdot \frac 1 { x ^ 2 } \, \mathrm d x = x \cdot f ( x ) - \int f ( x ) \, \mathrm d x = \ln x + C _ 1 \text ; $$ $$ \implies x \cdot \frac { f ( 1 ) } x - \int \frac { f ( 1 ) } x \, \mathrm d x = \ln x + C _ 1 \text ; $$ $$ \implies f ( 1 ) - f ( 1 ) \left ( \ln x + C _ 2 \right ) = \ln x + C _ 1 \text ; $$ $$ \implies \bigl ( f ( 1 ) + 1 \bigr ) \ln x = f ( 1 ) ( 1 - C _ 2 ) - C _ 1 \text . $$

The right side is a constant. The left side must also be a constant. $$ \implies f ( 1 ) + 1 = 0 \implies f ( 1 ) = - 1 \text . $$

Finally, we get $$ f ( x ) = - \frac 1 x \text ; $$ $$ \int \frac 1 { x ^ 2 } \, \mathrm d x = - \frac 1 x + C \text . $$

Remark 1.

My goal was to find antiderivative of $ \frac 1 { x ^ 2 } $. Because, $$ \int \frac 1 { x ^ 2 } \, \mathrm d x = \text {antiderivative of } \frac 1 { x ^ 2 } + \text{constant.} $$ In short, the function $ f ( x ) $ was an antiderivative.

Remark 2.

Initially $ a $ is a constant. However, it covers all real numbers except zero. Therefore, I replaced it with the variable $ x $.

How much of my work does math allow? Is the method I am using a nonsense? Otherwise, can the errors it contains be corrected?

Answer 1

Task

To prove that if $f :(0,\infty) \to \mathbb R$ is a function which satisfies $f(1) = -1$ and for every $x,y>0$, $f(y)-f(x) = \int_{x}^y \frac{dt}{t^2}$, then $f(x) = \frac{-1}{x}$ for all $x$. We use monotonicity of the integral and the change of variables formula, but do not use FTC.

Idea

• Create a functional equation for $f$ using a change of variables.

• Solve the functional equation.

The functional equation

Let $x,y,b>0$ be arbitrary. Then, consider $f(y)-f(x) = \int_x^y \frac{dt}{t^2}$. Perform the change of variables $u=bt$, then $du = bdt$. This results in : $$ f(y)-f(x) = \int_x^y \frac{dt}{t^2} = \int_{bx}^{by} \frac{b^2}{u^2}\frac{du}{b} = b\int_{bx}^{by} \frac{du}{u^2} = b(f(by)-f(bx)) $$

which leads to the equation : for all $b,x,y>0$, we have : $$ f(y)-f(x) = b(f(by)-f(bx)) $$

Solving the equation

As Professor Vector shows here, there exists a constant $a$ such that $xf(x)=ax-a-1$ for all $x>0$. Note that no tools from calculus were used. That is, $f(x) = a - \frac{a-1}{x}$.

But then, $\lim_{x \to \infty} f(x) = a$. Therefore, $\lim_{x \to \infty} (f(x) - f(1)) = a+1$ i.e. $a = \int_{1}^\infty \frac{dt}{t^2} - 1$. We expect $\int_1^\infty \frac{dt}{t^2} = 1$.

But how do we show that? The most obvious substitution $u = \frac {-1}t$ is barred since we are not allowed to use the fact that this is the antiderivative of $\frac{1}{t^2}$, which we are trying to prove! Furthermore, if we attempt to express this as a Riemann sum and take the limit, we could have done the same for general $n$ instead of $\infty$ and landed with the expression of $\frac{-1}{x}$ earlier.

Conclusion

While our approach eventually turned into a roundabout, we showed that $f(x) = a - \frac{a-1}{x}$ for all $x$, where $a = \int_1^{\infty} \frac{dt}{t^2}$, a quantity that unfortunately cannot be calculated without going back to the theory of the integral itself.

However, our sojourn lead us to a few important observations about such $f$, which is all we can do without the power formula.

• $0<\lim_{x\to \infty} \frac {dt}{t^2} = a < \infty$. The assertion that the integral is finite is quite surprising to those who have not seen the power law.

• $f(x)$ decays exactly at the rate $\frac 1x$ as $x \to \infty$ i.e. $\lim_{x \to \infty} \frac{f(x)-a}{x} = a-1$. This is because $f$ is a strictly increasing function, so we must have $a-1<0$ i.e. $a<1$ so $a-1 \neq 0$.

Answer 2

It's best to use definite integrals; this not only distinguishes certain variables your reasoning risks conflating (though you've done your best to explain why the steps are legal), but also makes it shorter. Define $f(x):=\int_\infty^x\frac{dt}{t^2}$ so $f(ax)=\int_\infty^{ax}\frac{dt}{t^2}=\int_\infty^x\frac{du}{au^2}=\frac{f(x)}{a}$; in particular, $f(a)=\frac{f(1)}{a}$. For $x>0$, integrate by parts with $u:=t,\,v:=f(t)$ so$$\begin{align}\ln x&=\int_1^x\tfrac{dt}{t}\\&=[tf(t)]_1^x-\int_1^xf(t)dt\\&=\underbrace{xf(x)-f(1)}_0-\int_1^x\tfrac{f(1)}{t}dt\\&=-f(1)\ln x\\\implies f(1)&=-1.\end{align}$$