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## Main Question or Discussion Point

Has anyone an idea how to start with this one?

[tex]

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

[/tex]

[tex]

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

[/tex]

- Thread starter dirk_mec1
- Start date

- #1

- 761

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Has anyone an idea how to start with this one?

[tex]

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

[/tex]

[tex]

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

[/tex]

- #2

Gib Z

Homework Helper

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- #3

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Thanks!Hello Dirk, and welcome to Physicsforums!

Okay here we go: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.

[tex] \int_0^{ \pi /4} u \cdot \tan(u)\ \mbox{d}u [/tex]

Which part, the u or the tangens?Integration by parts is the way to go for that new one.

- #4

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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".

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".

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- #5

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You're right! Are at least the boundaries correctly calculated?I think the integrand should be u*cot(u), since the x is in the denominator.

That's clear.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).

Never heard of Lipet before!Use LIPET (log, inverse trig, polynomial, exponential, trig function) to determine which function should be "u" when you integrate by parts.

Since we have a polynomial function, let that be "u" and let the cot function be "dv".

So we get:

[tex] \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 [/tex]

Assuming this correct how do I proceed?

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- #6

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

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

Gib Z

Homework Helper

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- #8

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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

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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

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

- #11

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dirk, what is the source of this problem?

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Just a buddy pointed out that this would be a 'challenge' (that's why it isn't in the homework section).dirk, what is the source of this problem?

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)

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oops, yes you are right.

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

- #17

Gib Z

Homework Helper

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- #18

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Sorry, an oversight. Should be Pi/4.

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What a mess! Why the answer isn't

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

why? Just curious.

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

why? Just curious.

- #20

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[tex]\frac{\text{Catalan}}{2}+\frac{\pi}{8}ln(2)[/tex]

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 [tex]\int_{0}^{1}\frac{tan^{-1}(x)}{x}dx=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)^{2}}=\text{Catalan}[/tex]

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

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

[tex]\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)[/tex]

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: [tex]\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k+1}}{2k+1}[/tex]

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

[tex]\int_{0}^{1}\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k}}{(x^{2}+1)(2k+1)}[/tex]

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.

[tex]\int_{0}^{1}\frac{x^{2k}}{x^{2}+1}\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2k+1}[/tex]

If we integrate [tex]\int x^{2k}dx[/tex], we get [tex]\frac{x^{2k+1}}{2k+1}[/tex].

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.

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