Integral Variation

flouran · 2009-07-19, 22:36

I was looking at an integral the other day:
$\int_2^{x^{1/3}}\frac{dt}{\log^3 t}$ .
And I was told that it varies as 1.5x/log x, because I was trying to show that:
$\frac{cx^{1/3}}{\log^3 x} \sim \frac{c}{\alpha}\int_2^{x^{1/3}}\frac{dt}{\log^3 t}$ , where $\alpha$ can be taken as 27.
I mean, I've taken BC Calc two years ago, I took Linear Algebra last year, and I will take Differential Equations in the coming fall, but I have never heard of the term "vary" when dealing with integrals, or perhaps I know of a term with a similar definition.
Would anyone mind explaining to me what it means that the integral varies as 1.5x/log x? (and/or give me a useful link talking about it).

cheesehead · 2009-07-20, 06:52

The variable ${t}$ is inside the integral. The variable ${x}$ is outside the integral.

Let

${f(x)} = \int_2^{x^{1/3}}\frac{dt}{\log^3 t}$ .

The claim is that

${f(x)}$ varies as 1.5x/log x.

Quote:

Originally Posted by flouran

I have never heard of the term "vary" when dealing with integrals, or perhaps I know of a term with a similar definition.

What's "varying" is the function ${f(x)}$ .

${f(x)}$ happens to have an integral inside it, but that's not important for "varies".

The function varies. The function has an integral inside it. Those are two separate statements. Combining them into "the integral varies" is just a contraction that happens to be okay because the function has only the integral in it (no other terms), so referring to the function as "the integral" is permissible here.

If the function were the sum of two different integrals (presumably both with an upper limit related to x), then one would need to say, "the sum of integrals varies as ..."

R.D. Silverman · 2009-07-20, 12:58

Quote:

Originally Posted by cheesehead

The variable ${t}$ is inside the integral. The variable ${x}$ is outside the integral.

Let

${f(x)} = \int_2^{x^{1/3}}\frac{dt}{\log^3 t}$ .

The claim is that

${f(x)}$ varies as 1.5x/log x.

[mathematical gibberish deleted]

If "varies" means "is asymptotic to", then the claim is wrong.

int 1/log^3 t dt = -t/(2 log t) - t/(2 log^2 t) + li(t)/2. := g(t)

Now evaluate g(x^1/3) - g(2).

flouran · 2009-07-20, 14:24

Quote:

Originally Posted by R.D. Silverman

int 1/log^3 t dt = -t/(2 log t) - t/(2 log^2 t) + li(t)/2. := g(t)

Now evaluate g(x^1/3) - g(2).

Thanks, that's what I wanted!

Primeinator · 2009-07-20, 15:57

Apparently I am rusty on my calculus as well. I have been through all the basic calculus classes (1-3) but I cannot seem to remember how to integrate that before you plug in your bounds. It is not possible to make a u-substitution either of u=log x or 1/log x... The only thing I can think of is integration by parts - tabular method but I do not see how this would work when you have one term that will never differentiate to zero. Dr. Silverman, how did you get this answer?

R.D. Silverman · 2009-07-20, 18:16

Quote:

Originally Posted by Primeinator

Apparently I am rusty on my calculus as well. I have been through all the basic calculus classes (1-3) but I cannot seem to remember how to integrate that before you plug in your bounds. It is not possible to make a u-substitution either of u=log x or 1/log x... The only thing I can think of is integration by parts - tabular method but I do not see how this would work when you have one term that will never differentiate to zero. Dr. Silverman, how did you get this answer?

Your thinking is correct......

Integration by parts combined with the definition of li(x).

[li(x) is not an elementary function]

Primeinator · 2009-07-20, 20:00

Quote:

Originally Posted by R.D. Silverman

Your thinking is correct......

Integration by parts combined with the definition of li(x).

[li(x) is not an elementary function]

Interesting. We never covered li(x) in Calc, but we did cover the sine integral in Ordinary Differential Equations (if you count doing one homework problem covering a topic). Looking it up, li(x) is dx/ln x whose domain depends upon the bounds of the integral. I assume that you let dv = 1/log x (easier to integrate) and u = 1/(log x)^2 when you did tabular integration by parts? However, I am not familiar with using this method when you have one term that will not differentiate to zero. How is it done when you have two transcendental functions? (A brief search on the internet was not very helpful regarding this).

I apologize, this is not my homework problem but I am always eager to learn something new

2009-07-19, 22:36	#1
flouran Dec 2008 7²×17 Posts	Integral Variation I was looking at an integral the other day: $\int_2^{x^{1/3}}\frac{dt}{\log^3 t}$ . And I was told that it varies as 1.5x/log x, because I was trying to show that: $\frac{cx^{1/3}}{\log^3 x} \sim \frac{c}{\alpha}\int_2^{x^{1/3}}\frac{dt}{\log^3 t}$ , where $\alpha$ can be taken as 27. I mean, I've taken BC Calc two years ago, I took Linear Algebra last year, and I will take Differential Equations in the coming fall, but I have never heard of the term "vary" when dealing with integrals, or perhaps I know of a term with a similar definition. Would anyone mind explaining to me what it means that the integral varies as 1.5x/log x? (and/or give me a useful link talking about it).

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2009-07-20, 15:57	#5
Primeinator "Kyle" Feb 2005 Somewhere near M52.. 7×131 Posts	Apparently I am rusty on my calculus as well. I have been through all the basic calculus classes (1-3) but I cannot seem to remember how to integrate that before you plug in your bounds. It is not possible to make a u-substitution either of u=log x or 1/log x... The only thing I can think of is integration by parts - tabular method but I do not see how this would work when you have one term that will never differentiate to zero. Dr. Silverman, how did you get this answer?