prove this integral does not exist

**szpengchao** · May 23rd 2008, 07:07 AM

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

**colby2152** · May 23rd 2008, 07:11 AM

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.

**arbolis** · May 23rd 2008, 08:12 AM

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.

**szpengchao** · May 23rd 2008, 08:14 AM

how can u get the first step

**Krizalid** · May 23rd 2008, 08:18 AM

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

**arbolis** · May 23rd 2008, 08:38 AM

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.

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