Help for the integration of a logarithm??

[KimDEJ]

I'm studying for my test on integrals and I've stumbled across something I find really confusing!

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\(\displaystyle \mbox{40e. }\, f(x)\, =\, {}^2\log\left(\dfrac{1}{x}\right)\, =\, {}^2\log\left(x^{-1}\right)\, =\, -^2\log(x)\)

. . . . .\(\displaystyle \Rightarrow\, F(x)\, =\, -\dfrac{1}{\ln(2)}\, \cdot\, \bigg(x\, \ln(x)\, -\, x\bigg)\, +\, c\, =\, \dfrac{-x\, \ln(x)\, +\, x}{\ln(2)}\, +\, c\)

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I understand that log2(1/X) = -log2(x), but I don't get why -log2(x) becomes -ln(x)/ln(2), instead of log(x)/log(2) and that you have to find the integral of ln(x) but not 1/ln(2). I can't find anything about this in my math book so I was wondering if someone could explain this to me?

 

[stapel]

I'm studying for my test on integrals and I've stumbled across something I find really confusing!

---

\(\displaystyle \mbox{40e. }\, f(x)\, =\, {}^2\log\left(\dfrac{1}{x}\right)\, =\, {}^2\log\left(x^{-1}\right)\, =\, -^2\log(x)\)

. . . . .\(\displaystyle \Rightarrow\, F(x)\, =\, -\dfrac{1}{\ln(2)}\, \cdot\, \bigg(x\, \ln(x)\, -\, x\bigg)\, +\, c\, =\, \dfrac{-x\, \ln(x)\, +\, x}{\ln(2)}\, +\, c\)

What is the definition of the "floating in the air" "2" preceding the "log" terms? What is the relationship between "f(x) and "F(x)"? What is the source of the "if-then" you've posted? Is this the answer to some (unstated) question? If so, what was that question? If not, what is it?

I understand that log2(1/X) = -log2(x), but I don't get why -log2(x) becomes -ln(x)/ln(2), instead of log(x)/log(2)...

It's the change-of-base formula (here). The choice of natural (base-e) log, rather than common (base-10) log is a matter of simplicity, since integrals are usually stated in terms of natural logs (and since they don't kick out extra multipliers).

 

[Steven G]

What is the definition of the "floating in the air" "2" preceding the "log" terms? What is the relationship between "f(x) and "F(x)"? What is the source of the "if-then" you've posted? Is this the answer to some (unstated) question? If so, what was that question? If not, what is it?


It's the change-of-base formula (here). The choice of natural (base-e) log, rather than common (base-10) log is a matter of simplicity, since integrals are usually stated in terms of natural logs (and since they don't kick out extra multipliers).

You don't know what the "floating in the air" "2" means? And I thought you knew just about everything.

 

[HallsofIvy]

I'm studying for my test on integrals and I've stumbled across something I find really confusing!

---

\(\displaystyle \mbox{40e. }\, f(x)\, =\, {}^2\log\left(\dfrac{1}{x}\right)\, =\, {}^2\log\left(x^{-1}\right)\, =\, -^2\log(x)\)

. . . . .\(\displaystyle \Rightarrow\, F(x)\, =\, -\dfrac{1}{\ln(2)}\, \cdot\, \bigg(x\, \ln(x)\, -\, x\bigg)\, +\, c\, =\, \dfrac{-x\, \ln(x)\, +\, x}{\ln(2)}\, +\, c\)

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I understand that log2(1/X) = -log2(x), but I don't get why -log2(x) becomes -ln(x)/ln(2), instead of log(x)/log(2) and that you have to find the integral of ln(x) but not 1/ln(2). I can't find anything about this in my math book so I was wondering if someone could explain this to me?

From the definition of \(\displaystyle log_2(x)\), "log base 2 of x", if \(\displaystyle y= log_2(x)\) then \(\displaystyle x= 2^y\). Take the natural logarithm of both sides:
\(\displaystyle ln(x)= ln(2^y)= yln(2)\). So \(\displaystyle y= log_2(x)= \frac{ln(x)}{ln(2)}\).

More generally, by the same argument, \(\displaystyle log_a(x)= \frac{log_b(x)}{log_b(a)}\) for any bases (positive numbers) a and b.

 

[mathguy11]

Integration by Parts

First, in order to see why log_2(x) = log(x)/log(2), you must think about what the definition of an logarithm is- the inverse of the exponent function. So, let's say 2^a = x, and if you take ln() of both sides, you get ln(2^a) = ln(x). By the exponent property of logarithms, you get a*ln(2) = ln(x), so a = ln(x)/ln(2). In addition, recall that the inverse of 2^a = x is log_2(x). So log_2(x) = ln(x)/ln(2).


In your problem, you're left with -ln(x)/ln(2), and you must integrate that. -1/ln(2) is just a coefficient (it equals -1.4426...). It's analogous to integrating 2x versus integrating x. You only have to integrate the actual function, which is ln(x).


Integrating ln(x) is fairly tricky, and it involves something that's called integration by parts. Here's my solution to the integral of ln(x): http://www.mathwizurd.com/calc/2016/2/16/integration-by-parts. (Sorry, it's too long to type out here) Basically, it's the product rule, but for integrals.


Anyway, hope this helped!