Prove the improper integral converges

[Hernaner28]

Homework Statement

(a). Prove that the improper integrals converge:

\displaystyle \int\limits_{0}^{1}{\frac{\ln x}{1+{{x}^{2}}}}dx

\displaystyle \int\limits_{1}^{\infty }{\frac{\ln x}{1+{{x}^{2}}}}dx

And relate each other.

(b) Deduce the value of:

\displaystyle \int\limits_{0}^{\infty }{\frac{\ln x}{1+{{x}^{2}}}}dx

Homework Equations

The Attempt at a Solution

I managed to prove that \displaystyle \int\limits_{0}^{1}{\frac{\ln x}{1+{{x}^{2}}}}dx converges by bounding it by \displaystyle \frac{1}{\sqrt{x}} which converges. Now, I cannot bound the other integral so I applied integration by parts and obtained:

\displaystyle \ln x\cdot \arctan x\left. \begin{align}<br /> &amp; \\ <br /> &amp; \\ <br /> \end{align} \right|_{1}^{\infty }-\int\limits_{1}^{\infty }{\frac{\arctan x}{x}dx}

But the term on the left tends to infinity. What can I do? And how do I relate them?

Thank you!

 

[cjc0117]

\frac{lnx}{1+x^{2}}≤\frac{lnx}{x^{2}} for x≥1

 

[Hernaner28]

Thank you! And how can I deduce the value without working it out?

 

[SammyS]

Homework Statement

(a). Prove that the improper integrals converge:
\displaystyle \int\limits_{0}^{1}{\frac{\ln x}{1+{{x}^{2}}}}dx
\displaystyle \int\limits_{1}^{\infty }{\frac{\ln x}{1+{{x}^{2}}}}dx
And relate each other.

(b) Deduce the value of:
\displaystyle \int\limits_{0}^{\infty }{\frac{\ln x}{1+{{x}^{2}}}}dx

Homework Equations

The Attempt at a Solution

I managed to prove that \displaystyle \int\limits_{0}^{1}{\frac{\ln x}{1+{{x}^{2}}}}dx converges by bounding it by \displaystyle \frac{1}{\sqrt{x}} which converges. Now, I cannot bound the other integral so I applied integration by parts and obtained:

\displaystyle \ln x\cdot \arctan x\left. \begin{align}<br /> &amp; \\ <br /> &amp; \\ <br /> \end{align} \right|_{1}^{\infty }-\int\limits_{1}^{\infty }{\frac{\arctan x}{x}dx}

But the term on the left tends to infinity. What can I do? And how do I relate them?

Thank you!

The integral \displaystyle \int_1^\infty\frac{dx}{\sqrt{x}} does not converge.

 

[Hernaner28]

It does NOT converge when x tends to infinity but it DOES converge when x tends to zero.

 

[algebrat]

What class is this for, which school.

 

[Hernaner28]

Sorry?

 

[algebrat]

It does NOT converge when x tends to infinity but it DOES converge when x tends to zero.

By "it", I believe Hernaner was referring to the bound

\displaystyle \int\frac{dx}{\sqrt{x}}

The integral 0\le\displaystyle \int_1^\infty\frac{\ln(x)}{1+x^2}dx&lt;\infty

 

[algebrat]

The integral 0\le\displaystyle \int_1^\infty\frac{\ln(x)}{1+x^2}dx&lt;\infty

I see this intuitively since ln(x) grows slower than x, so ln(x)/(1+x^2) grows slower than x/x^2=1/x.

 

[cjc0117]

The problem now it seems is figuring out how the two integrals relate to each other. I know that \int^{1}_{0}\frac{lnx}{1+x^{2}}dx=-\int^{∞}_{1}\frac{lnx}{1+x^{2}}dx but I can't figure out how to prove it. OP, can you answer algebrat's questions? What class is this for?

 

[Hernaner28]

How to do you know that the absolute value of both integrals are the same?

What do you mean by "Class"? This is an exercise from my Calculus 1 course

Thanks!

 

[cjc0117]

I know because I cheated using wolframalpha. And by class I meant course, which you just said was calculus 1. Wolfram alpha says something about Catalan's constant. Maybe you should look into that.

 

[Hernaner28]

No idea about Catalan's constant! hehe

Thanks

 

[algebrat]

This is an exercise from my Calculus 1 course

If it were an engineering class, one might be more welcome to calculate with a computer.

When I took calc 1, I don't remember seeing problems like prove this improper integral exists. Is calc 1 the class where you first learn about derivatives?

As for getting the value, I can't see how to do it with any methods, are you sure there is not a special note on this from your texbook or lecture notes.

Is this a homework problem from the back of the book or is this like a research project?

 

[cjc0117]

I figured it out. Starting with \int^{1}_{0}\frac{lnx}{1+x^{2}}dx, use the change of variable u=1/x.

 

[algebrat]

I figured it out. Starting with \int^{1}_{0}\frac{lnx}{1+x^{2}}dx, use the change of variable u=1/x.

Nice cjc0117. Perhaps the moral of the story is after you plug it into wolfram alpha, you find they're equal and opposite, so you try to show that. My first thought was it was symmetric about some line, but when that didn't work, I dropped that lead. I wish my brain was big enough to remember all possible leads. Somehow the limits 0..1 and 1..∞ might have suggested various substitutions, but that would have taken a certain type of mental energy that cjc0117 did have.

That's a very hard problem, especially without a computer. I'm surprised it's in calc 1.

 

[Hernaner28]

Thank you! I will write that down!

In my Calc course each week teachers upload a sheet with exercises according to the week topic.

Thanks!

 

[cjc0117]

algebrat, I thought it might have had something to do with symmetry as well. Admittedly, I didn't figure it out all by myself. I used wolfram alpha to find out they're equal but opposite, and that the integral is equal to Catalan's constant, then I looked up this article about Catalan's constant:

https://docs.google.com/a/husky.neu...&sig=AHIEtbR6mFROrq-fZzBIANxlgJznBBYVBg&pli=1

So I did the problem in reverse basically. I don't think I would have guessed to use that substitution right off the bat.

 

[algebrat]

algebrat, I thought it might have had something to do with symmetry as well. Admittedly, I didn't figure it out all by myself. I used wolfram alpha to find out they're equal but opposite, and that the integral is equal to Catalan's constant, then I looked up this article about Catalan's constant:

https://docs.google.com/a/husky.neu...&sig=AHIEtbR6mFROrq-fZzBIANxlgJznBBYVBg&pli=1

So I did the problem in reverse basically. I don't think I would have guessed to use that substitution right off the bat.

As I progressed through my undergraduate career (especially graduate algebra), I found it harder to complete assignments without a little "research" help. Sometimes I would only read a source till I saw a hint and would work out the rest, and I always cited what hints I used (even saying things like, "I only read the first sentence"). I think there is an art to it, the digging can get as intricate as yours was, going from knowledge of convergence, finding the value zero on a computer, following the Catalan lead, to finding some particular document where they suggest a transformation. While it is maybe not completely in the spirit of academia, it is totally the sort of thing you need a little of in a work setting (still need independent imagination too), and I would guess even in academia, there is a fair amount of putting things together from other's knowledge.

(It's the strange problems like these that fall a little farther from the foundations (or harder to see the path of solution) that we learn in the material that seem to warrant this sort of research.)

I was never happy with my tendency to lean on the internet in graduate algebra (almost once or twice per homework), so a side project of mine is to simplify the subject (as it sits in my head) so that I get it more.

Hernaner28, keep in mind that there is a possibility that your instructor has some example in notes where they recently did a certain manipulation. That would be a more academically accepted form of research, where your instructor just recently did a similar manipulation. But I couldn't bet a dollar on it, just an idea.

 

[SammyS]

I figured it out. Starting with \int^{1}_{0}\frac{lnx}{1+x^{2}}dx, use the change of variable u=1/x.

OK: After doing the substitution, I see what you're getting at.

u = 1/x

du = -1/x² dx

\displaystyle \int_1^t{\frac{\ln(x)}{1+{{x}^{2}}}}dx\quad\to \quad\int_1^{1/t}{\frac{-\ln(u) }{-(u^2)(1+{1/{u}^{2}})}}du

So that \displaystyle \lim_{t\to\infty}\int_1^t{\frac{\ln(x)}{1+{{x}^{2}}}}dx=\lim_{t\to\infty}-\int_{1/t}^1{\frac{\ln(u)}{u^2+1}}du

\displaystyle \int_1^\infty{\frac{\ln(x)}{1+{{x}^{2}}}}\,dx=-\int_{0}^1{\frac{\ln(x)}{1+x^2}}\,dx

Added in Edit:
Lower limit of integration corrected per following post.

 

[cjc0117]

Yeah, exactly.

EDIT: except for the very last equation. The lower limit is 1, not 0. But I think you know that.

 

[Ray Vickson]

Homework Statement

(a). Prove that the improper integrals converge:

\displaystyle \int\limits_{0}^{1}{\frac{\ln x}{1+{{x}^{2}}}}dx

\displaystyle \int\limits_{1}^{\infty }{\frac{\ln x}{1+{{x}^{2}}}}dx

And relate each other.

(b) Deduce the value of:

\displaystyle \int\limits_{0}^{\infty }{\frac{\ln x}{1+{{x}^{2}}}}dx

Homework Equations

The Attempt at a Solution

I managed to prove that \displaystyle \int\limits_{0}^{1}{\frac{\ln x}{1+{{x}^{2}}}}dx converges by bounding it by \displaystyle \frac{1}{\sqrt{x}} which converges. Now, I cannot bound the other integral so I applied integration by parts and obtained:

\displaystyle \ln x\cdot \arctan x\left. \begin{align}<br /> &amp; \\ <br /> &amp; \\ <br /> \end{align} \right|_{1}^{\infty }-\int\limits_{1}^{\infty }{\frac{\arctan x}{x}dx}

But the term on the left tends to infinity. What can I do? And how do I relate them?

Thank you!

If
A = \int_0^1 \frac{\ln(x)}{1+x^2}\, dx, you can analyze the convergence of A by looking at A_1 = \int_0^1 \ln(x)\,dx and A_2 = \int_0^1 g(x)\,dx, where
g(x) =\frac{\ln(x)}{1+x^2} - \ln(x) = - x^2 \ln(x) + x^4 \ln(x) + O(x^6). Clearly, A₂ is finite, so you need only look at A₁, which you can evaluate explicitly.

By an appropriate change of variable you ought to be able to relate the integral from 1 to ∞ to the integral from 0 to 1.

RGV