# Thread: prove this integral does not exist

1. ## prove this integral does not exist

integral ( 1/x, x, 0,1)

2. Originally Posted by szpengchao
integral ( 1/x, x, 0,1)
Is that $\int_0^1 \frac{1}{x} dx$?

If so, note it will not exist at the bound: $x=0$ because not only is the function not continuous at that point (cannot divide by zero), the resultant integral $\ln(x)$ cannot take values less than or equal to zero.

3. You can notice that $\int_0^1 \frac{1}{x} dx=\int_1^\infty \frac{1}{x} dx$.
So we have $\int_1^\infty \frac{1}{x} dx=$ limit when b tends to $\infty$ of $\int_1^b \frac{1}{x}dx =$ limit when b tends to $\infty$ of $ln(x)$ evaluated in $b$ minus $ln(x)$ evaluated in $1$. Which is equal to limit when b tends to $\infty$ of $ln(x)$ evaluated in $b$, which is $\infty$, thus divergent.

4. ## prove

how can u get the first step

5. It is a simple reciprocal substitution, try it with $x=\frac1u.$

6. Ok. I said $\int_0^1 \frac{1}{x} dx=\int_1^\infty \frac{1}{x} dx$, this implies that the graph of $f(x)=\frac{1}{x}$ is symmetric with respect to the graph of $y=x$. It means that if $f(x)=\frac{1}{x}$, then $f^{-1}(x)=f(x)$. In other words, f has its own bijection equals to itself.
Now to prove it, you just have to show that $f(f^{-1}(x))=x$.

$f(f^1(x))$ is equals to $\frac{1}{f^1(x)}=\frac{\frac{1}{1}}{\frac{1}{x}}=x$. It is demonstrated. If we take a precise example maybe you'll understand better. If $x=5$, then $f(5)=\frac{1}{5}$. That means that to get 5 with our function, we need to evaluated it in $\frac{1}{5}$, which works. Maybe you will understand even better seeing the graph. It's trivial that the area under the curve of $f(x)=\frac{1}{x}$ from $0$ to $1$ is equal to the area from 1 to $\infty$. To be rigorous, just say f is its own bijection (and show it as I did for you), that should be enough.