What is the Integral of f(x)=1/x from x=1 to x=infinite?

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The integral of f(x) = 1/x from x=1 to x=infinity is evaluated using the limit approach, resulting in the expression ∫_1^∞ (1/x) dx = lim (t→∞) ∫_1^t (1/x) dx. The antiderivative of 1/x is ln|x| + C, derived from the fact that the derivative of ln(x) is 1/x. However, the integral diverges, leading to the conclusion that ∫_1^∞ (1/x) dx is undefined or infinite. The discussion clarifies the connection between the integral and the natural logarithm function, emphasizing the importance of understanding limits in improper integrals. Thus, the integral does not yield a finite numerical answer.
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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?
 
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\int \frac{1}{u}\,du=\ln{|u|}+C
 
The derivative of ln x is 1/x so like the person stated above

\int (1/u)du = ln|u| + C
 
I'm sorry, but I'm so lost already. Could somebody please explain it to me? What is u?
 
Aki said:
I'm sorry, but I'm so lost already. Could somebody please explain it to me? What is u?
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, that's all you need to know.
 
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
 
kreil said:
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

I never learned that yet. That's pretty cool.
 
so basically, there's not "number" answer to that questions? The answer is just a function?
and also where did ln(x) come from?
 
\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 \frac{d}{dx} lnx = 1/x [/tex]... so your next question is why this is true...<br /> <br /> assume you know product rule and the derivative of e^x is e^x itself<br /> <br /> e^{\ln{x}} = x<br /> <br /> \frac{d}{dx} e^{\ln{x}} = \frac{d}{dx} x<br /> <br /> e^{\ln{x}} \frac{d}{dx} (\ln{x}) =1 --------product rule<br /> <br /> x \frac{d}{dx} (\ln{x})=1 <br /> <br /> \frac{d}{dx} (\ln{x}) = \frac{1}{x}<br /> <br /> so the anti-derivative of 1/x is ln(x)
 
  • #10
Erm vincentchan don't you mean the chain rule?
 
  • #11
Yes,it is the chain rule...Anyway,the result is correct and the method of finding it is correct as well...

Daniel.
 

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