# Function f(x)=1/x intergral

1. Jan 25, 2005

### Aki

I have the function f(x)=1/x , and there's two asymptotes, the x-and y-axes. As x gets larger, the f(x) becomes smaller, but it is never 0. So my question is, what is the intergral from x=1 to x=infinite?

2. Jan 25, 2005

### Sirus

$$\int \frac{1}{u}\,du=\ln{|u|}+C$$

3. Jan 25, 2005

### digink

The derivative of ln x is 1/x so like the person stated above

$$\int (1/u)du = ln|u| + C$$

4. Jan 25, 2005

### Aki

I'm sorry, but I'm so lost already. Could somebody please explain it to me? What is u?

5. Jan 25, 2005

### digink

Let me put it in a simpler form, just replace u with x something you are probably more common to seeing.

Now the derivative for the ln x = 1/x.

now if you have $$\int {1/x}dx$$ you know that's the derivative of the ln of x, so you end up with that = $$ln|x| + C$$

this is just based of knowing the derivative and antiderivative of ln x, thats all you need to know.

6. Jan 25, 2005

### kreil

integrals at infinity are calculated by

$$\int_1^{\infty}f(x)dx=\lim_{t{\rightarrow}\infty}\int_1^tf(x)dx$$

if you use this in combination with the info above you can calculate it

7. Jan 25, 2005

### JasonRox

I never learned that yet. That's pretty cool.

8. Jan 28, 2005

### Aki

so basically, there's not "number" answer to that questions? The answer is just a function?
and also where did ln(x) come from?

9. Jan 28, 2005

### vincentchan

$$\int_1^{\infty} \frac{1}{x}dx$$ is undefined, or infinite.. depend on which one you feel more comfortable

where did ln x came from...hmmm... it came from [itex] \frac{d}{dx} lnx = 1/x [/tex].... so your next question is why this is true.....

assume you know product rule and the derivative of e^x is e^x itself

$$e^{\ln{x}} = x$$

$$\frac{d}{dx} e^{\ln{x}} = \frac{d}{dx} x$$

$$e^{\ln{x}} \frac{d}{dx} (\ln{x}) =1$$ --------product rule

$$x \frac{d}{dx} (\ln{x})=1$$

$$\frac{d}{dx} (\ln{x}) = \frac{1}{x}$$

so the anti-derivative of 1/x is ln(x)

10. Jan 28, 2005

### Zurtex

Erm vincentchan don't you mean the chain rule?

11. Jan 28, 2005

### dextercioby

Yes,it is the chain rule...Anyway,the result is correct and the method of finding it is correct as well...

Daniel.