What is the integral of tangent x divided by x from 0 to infinity?

[dextercioby]

Homework Statement

Compute \int_{0}^{\infty} \frac{\tan x}{x} {} dx

Homework Equations

The Attempt at a Solution

Got no idea, obviously the derivation differs from the \int_{-\infty}^{\infty} \frac{\sin x}{x} {} dx one. The result I'm supposed to get is \frac{\pi}{2}.

 

[Yura]

i would do integration by parts on the 1/x and the tan(x) then work from there.

probably let tan(x) equal sin(x)/cos(x) and use substitution integration to solve that part

then when you've solved the integration, sub in infinity and just logic it out from there i guess.

sorry if its not much help, i can't think straight at the moment. (halfway thought working on my own integration problem)

 

[Gib Z]

No point, that integral is not expressible in terms of elementary functions. Must be some property with the bounds thatll help then, rewrite as \lim_{a\to\infty} \int^a_0 \frac{\tan x}{x} dx and try u= a - x

 

[quantumdude]

I'm surprised that the interval even converges, since the integrand is so badly behaved. But if it is supposed to come out to \pi/2, then I suppose you can do a numerical approximation using the Maclaurin series for sin(x) and cos(x).

 

[VietDao29]

Errr... Why do I have a feeling that this integral does not converge. >__<

 

[Dick]

Errr... Why do I have a feeling that this integral does not converge. >__<

Probably because in the strictest sense, it doesn't. It's an improper integral. And so treating it numerically would be a disaster. It may have a shadowy existence via a contour integral argument.

 

[Gib Z]

Maclaurin/Taylor series approach does not work, tried very hard.

 

[Data]

As Dick alluded to, if the integral has a value, the argument's going to have to use complex analysis, as for \int_{0}^\infty \frac{\sin x}{x} dx. Of course, this at least appears to be a lot more complicated thanks to the integrand's infinitely many poles on the real line. For example, the sum of the residues on the positive real line diverges! Maple seems to agree with the \frac{\pi}{2} value, though.

 

[Gib Z]

You know its not a good sign when dexter has a problem with an integral ...God Help Us!

 

[dextercioby]

So nobody could help me, right ?

 

[Gib Z]

I did some numerical analysis of the graph of the function in question, and i found that at all the points the function is equal to zero, k pi, the area from kpi - 1 and kpi +1, is always negative and decreasing. I chose a difference of 1, but any value less than pi/2 will work.

So All I can say is that the integral \int_{\frac{\pi}{2}}^{\infty} \frac{\tan x}{x} dx is less than zero, and very badly approximated to be -3.775239041.

I also approximated the area from 0 to 1.57 ( I couldn't do exactly pi/2 for obvious reasons) is 5.5861702.

So your integral should be the addition of the two integrals, and I got 1.810931159. I looks like its slowly getting to pi/2, but what I just did was completely useless. O well, I tried.

 

[Dick]

So nobody could help me, right ?

Did you try defining it as a contour integral? Data said that the sum of the poles on the real line diverged. Just looking at the integral I would guess they would actually form an alternating series.

 

[Gib Z]

Just looking at the integral I would guess they would actually form an alternating series.

Yup It does :) I haven't done Contour Integration yet though, So I can't tell you the specifics of the series, just that its alternating and converges to \frac{\pi}{2} :) Sorry dex.

 

[Dick]

Yup It does :) I haven't done Contour Integration yet though, So I can't tell you the specifics of the series, just that its alternating and converges to \frac{\pi}{2} :) Sorry dex.

Well, I'm doubting that now. But if it does, then fill us in. Because I'm currently stumped. It's funny if integral of sin(x)/x and tan(x)/x are equal 0 to infinity.

 

[Gib Z]

Well this is kind of hard for me to explain but here's what I saw.

At every point k\cdot \pi, k\in \mathbb{Z}, f(x) is zero as we know. The integral from k\pi - z, k\pi + z, z &lt; \pi/2 is always negative, and decreases as k increases. So What I meant was that the areas from k*pi -x, to kpi, summed with k*pi + x, produce an alternating series.

 

[dextercioby]

I can't see any possible contour that would allow me to compute the integral.

 

[Dick]

I don't either - nor can I see a way to manipulate Si(x). How do you know the integral is pi/2? Where did the problem come from?

 

[dextercioby]

It came from my mind, i just fooled around with Maple and it gave pi/2. I looked up the integral in the bible G & R and the value of pi/2 was confirmed.

 

[Dick]

I would post a query on the sci.math newsgroup. There's a lot giant brains over there.

 

[zoki85]

Sorry to burst your bubble but

Homework Statement

Compute \int_{0}^{\infty} \frac{\tan x}{x} {} dx

Homework Equations

The Attempt at a Solution

Got no idea, obviously the derivation differs from the \int_{-\infty}^{\infty} \frac{\sin x}{x} {} dx one. The result I'm supposed to get is \frac{\pi}{2}.

This integral ,the way you wrote it,is said to be DIVERGENT.
That's why nobody could help you.
Another ,more trivial, example is:
\int_{\pi}^{2\pi}\tan x dx
This is divergent too.

 

[Gib Z]

Ahh no it isn't divergent. Just because it is an improper integral doesn't mean it diverges. And the second integral converges as well, its 0.

 

[Dick]

I posted this to the sci.math group and after a number of replies saying such things as i) that can't be right, you are confusing it with sin(x)/x and ii) that can't be in G&R. I finally got the following. It sounds promising. I'll have to try it sometime.

OK. Think of integrating tan(z)/z around a rectangle, where real
part goes from -M*pi to M*pi, imaginary part goes from 0 to R.
At the simple poles (2*k+1)Pi/2, use the principal value.
[These residues cancel in +/- pairs.] Fixing R, when
M goes to infinity (along integers), the integrals on the two ends
go to zero (because of the denominator), so the real integral
we are interested in is the same as the integral along the horizontal
line x+i*R, where R is large and positive. But tan(x+i*R)
goes uniformly to i as R -> infinity, so this upper integral converges
to int(i/(x+i*R), x=-infinity..infinity), and that is, indeed, pi
(in the principal value sense, the limit of the integral -M to M).

Our integral from -infinity to infinity, then, is pi, so
our integral from 0 to infinity is pi/2.

--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/

 

[zoki85]

Ahh no it isn't divergent. Just because it is an improper integral doesn't mean it diverges. And the second integral converges as well, its 0.

Ahh YES,integral \int_{\pi}^{2\pi}\tan x \, dx diverges my friend.

Similarily,and better example for that matter, is integral:
\int_{-1}^{+1}\frac{dx}{x}
which diverges as well.Every serious mathematician will tell you that.
You may say it's 0,but I may argue that it's 1000000.Who's right?
You can't make sense of (-\infty + \infty) just like that.
What may approach 0 and make sense is specification of the limit:

\lim_{b\to 0+}(\int_{-1}^{-b}\frac{dx}{x}+\int_{b}^{1}\frac{dx}{x}) which doesn't have to be equivalent with \int_{-1}^{+1}\frac{dx}{x}.

Math doesn't tolerate sloppiness,specially not when working with singularities and infinities.

 

[Gib Z]

I can not say I am a serious mathematician, but I'm sure I know what I'm talking about. Just because the function diverges in the domain of integration, does not mean the integral does.

I know that the anti derivative of tan x is (-\log_e |\cos x|) and The logical substitutions and subtractions yeild zero.

Have you ever tried looking at a graph of tan x between pi and 2pi? Notice the symmetry? The integrals from pi to 3pi/2, and 3pi/2 to 2pi cancel out!

Same goes for 1/x, and its zero again!

 

[zoki85]

You appear like not to be reading my post.
I know what you are trying to say ,and that there's obvious symmetry in graph of the both functions.That's why I given the examples !
My point to you is just to do this:ask any professor of mathematics if improper integral \int_{-1}^{+1}\frac{dx}{x} diverges.
You will see what his/her answer is!

 

[Gib Z]

Is this a joke? I am reading your post very carefully and I know that its not true. I know that I am right.

 

[zoki85]

Is this a joke? I am reading your post very carefully and I know that its not true. I know that I am right.

No,it's not joke.
I told you what to do.

 

[Gib Z]

Ask a math professor? Calculus is my strongest subject, I could teach University Students this stuff and I'm 15 years old. Can someone else reading this help me out? I am 100% certain that I am correct.

 

[HallsofIvy]

Gib Z, the definition of an improper integral, such as
\int_{-1}^1 \frac{1}{x}dx
is
\lim_{\alpha\rightarrow 0^-}\int_{-1}^{\alpha} \frac{1}{x}dx+ \lim_{\beta\rightarrow 0^+} \int_{\beta}^1 \frac{1}{x}dx
and the two limits must be taken independently. While I can't speak for tan x/x, The integral of 1/x is divergent over any interval containing 0.

What you are referring to. using symmetry so that you are effectively doing
\lim_{\alpha\rightarrow 0^-}\int_{-1}^{\alpha} \frac{1}{x}dx+ \lim_{\alphja\rightarrow 0^+} \int_{\alpha}^1 \frac{1}{x}dx
is the "Cauchy Principal Value" which can be used in some circumstances where the integral itself does not converge but is NOT the integral itself.

 

[Gib Z]

Thank you Halls, I know that. But look at zoki85's post, he's asking for the Cauchy Principle Value!