# Prove the improper integral converges

1. Jun 17, 2012

### Hernaner28

1. The problem statement, all variables and given/known data
(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:

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

2. Relevant equations

3. 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} & \\ & \\ \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!

2. Jun 17, 2012

### cjc0117

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

3. Jun 17, 2012

### Hernaner28

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

4. Jun 17, 2012

### SammyS

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

5. Jun 17, 2012

### Hernaner28

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

6. Jun 17, 2012

### algebrat

What class is this for, which school.

7. Jun 17, 2012

### Hernaner28

Sorry?

8. Jun 17, 2012

### algebrat

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

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

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

9. Jun 17, 2012

### algebrat

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.

10. Jun 17, 2012

### cjc0117

The problem now it seems is figuring out how the two integrals relate to eachother. 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?

11. Jun 17, 2012

### 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!

12. Jun 17, 2012

### 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.

13. Jun 17, 2012

### Hernaner28

No idea about Catalan's constant! hehe

Thanks

14. Jun 17, 2012

### algebrat

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?

15. Jun 17, 2012

### cjc0117

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

16. Jun 17, 2012

### algebrat

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.

17. Jun 17, 2012

### 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!

18. Jun 17, 2012

### 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:

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

19. Jun 17, 2012

### algebrat

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.

Last edited: Jun 17, 2012
20. Jun 17, 2012

### SammyS

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

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