# Difficult integral

1. May 28, 2008

### dirk_mec1

$$\int_0^1\ \frac{\arctan(x)}{x(x^2+1)}\ \mbox{d}x$$

2. May 28, 2008

### Gib Z

Hello Dirk, and welcome to Physicsforums! As for that integral, a good step might be to let arctan x = u, then it makes the integral look much more approachable, you might have even seen it before. Integration by parts is the way to go for that new one.

3. May 28, 2008

### dirk_mec1

Thanks!

Okay here we go:

$$\int_0^{ \pi /4} u \cdot \tan(u)\ \mbox{d}u$$

Which part, the u or the tangens?

4. May 28, 2008

### snipez90

I think the integrand should be u*cot(u), since the x is in the denominator. Also, if you didn't know the derivative of arctan(x), the substitution x = tan(t) (or u) works just as well provided you know trig (it's essentially the same substitution).

I use LIPET (log, inverse trig, polynomial, exponential, trig function) to determine which function should be "u" when you integrate by parts. The natural logarithm takes precedence, and then so on. Since we have a polynomial function, let that be "u" and let the cot function be "dv".

Last edited: May 28, 2008
5. May 28, 2008

### dirk_mec1

You're right! Are at least the boundaries correctly calculated?

That's clear.

Never heard of Lipet before!

So we get:

$$\int_0^{\pi /4} u \cdot \cot(u)\ \mbox{d}u = u \cdot \ln( \vert \sin(u) | ) |_{0}^{ \pi /4} - \int_{0}^{\pi /4} \ln(| \sin(u) | )\ \mbox{d}u$$

Assuming this correct how do I proceed?

Last edited: May 28, 2008
6. May 28, 2008

### snipez90

Yes, the limits of integration are correct. I'm trying to work on the rightmost integral (since that'll solve the problem), but I can't find an easy trick. I know that

$$\int_{0}^{\pi /2} \ln( \sin(u) )\ \mbox{d}u = -{\pi}\ln(2)/2$$

because I computed it yesterday, but with the different upper limit of integration, this integral seems even harder. I'll keep trying, perhaps I've overlooked a nice substitution.

7. May 29, 2008

### Gib Z

Ahh Sorry guys, my very bad mistake :( It seems x cot x has no elementary derivative >.< And the original substitution doesn't have any obvious substitution to me, It must be something nice with the bounds. Sorry again!

8. May 29, 2008

### dirk_mec1

Great, what do i have to do now? Series expansion? I don't think there's a better substitution than u=arctan(x), right?

9. May 29, 2008

### exk

I am just curious, are you sure that there is a closed form solution to this integral?

I have been playing around with it and no matter what substitution I picked I got integrals that are inexpressible in terms of elementary functions (the other substitution combined with change of variable produces x*tan(x)dx).

Perhaps a numeric integration method might work for this.

10. May 29, 2008

### dirk_mec1

I think you're right, this integral can not be expressed in elementary functions.

11. May 29, 2008

### snipez90

dirk, what is the source of this problem?

12. May 30, 2008

### dirk_mec1

Just a buddy pointed out that this would be a 'challenge' (that's why it isn't in the homework section).

I managed to find some info about it but I understand it only to a certain level:

h ttp://mathworld.wolfram.com/AhmedsIntegral.html (remove the space in the beginning)

13. May 30, 2008

### exk

I am not sure that the integral you linked is related to the problem your friend gave. Where did your friend find it?

14. May 30, 2008

### dirk_mec1

Look better! The exact integral is present in the link!

15. May 30, 2008

### exk

oops, yes you are right.

16. Jan 8, 2009

Deduce the area under the curve= p x cot p from 0 to pi/4.

17. Jan 8, 2009

### Gib Z

Perhaps you should have read through the entire thread before posting this. If the upper bound were pi/2, we could have solved it as well.

18. Jan 8, 2009

Sorry, an oversight. Should be Pi/4.

19. Jan 9, 2009

What a mess! Why the answer isn't

ln sin pi/4 {pi/4 - sin pi/4} + 1,

why? Just curious.

20. Jan 9, 2009

I was pondering this integral a little today. I see from the MathWorld site that the solution is

$$\frac{\text{Catalan}}{2}+\frac{\pi}{8}ln(2)$$

I think if we are clever enough, we may be able to transform the integral into some other integrals that can be done. Though not by elementary means.

It can be shown that $$\int_{0}^{1}\frac{tan^{-1}(x)}{x}dx=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)^{2}}=\text{Catalan}$$

Also, it can be shown that $$\int_{0}^{1}\frac{ln(x+1)}{x^{2}+1}dx=\frac{\pi}{8}ln(2)$$

Can we do some manipulations and transform our integral into this:

$$\frac{1}{2}\int_{0}^{1}\frac{tan^{-1}(x)}{x}dx+\int_{0}^{1}\frac{ln(x+1)}{x^{2}+1}dx=\frac{K}{2}+\frac{\pi}{8}ln(2)$$

I will have to look at it some more tomorrow. Fun and challenging problem.

I am certain a lot of you out there are certainly better than me at this, so what do you think?.

It is there. I can smell it. I see many different approaches and identities, but am missing something.

I also tried the series for arctan: $$\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k+1}}{2k+1}$$

If we use that with the rest of our integral, we get:

$$\int_{0}^{1}\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k}}{(x^{2}+1)(2k+1)}$$

Now, some manipulations here and there and it looks like it should fall into place.

There is uniform convergence from 0 to 1, so we can switch our signs around.

$$\int_{0}^{1}\frac{x^{2k}}{x^{2}+1}\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2k+1}$$

If we integrate $$\int x^{2k}dx$$, we get $$\frac{x^{2k+1}}{2k+1}$$.

Which when multiplied with the other 2k+1, we get the catalan constant.

Like I said, it is right there, but..............:uhh:

I am just throwing some things out there. Having fun with the problem.