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!

 

[Dick]

I've just realized that I have been using the phrase 'improper integral' wrong - that's an integral that is convergent but unbounded. What I meant was an integral that has a 'principal value' - as Halls defined it but is not convergent. Sorry if this caused any confusion. As Dr. Edgar's recipe says, the sense in which a number can be applyed to this integral is exactly in the principal value sense.

 

[Geekster]

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}.

Well, I, couldn't work it out.

But I found that maple can work out the indefinite integral, and the definite integral which as you know is Pi/2.

While I don't know how to derive the indefinite integral, with your limits with the following does work out quite nicely.

<br /> -i \left( \ln \left( x \right) -2\,\int \!{\frac {1}{x \left( {e^{2\,<br /> <br /> ix}}+1 \right) }}{dx} \right)

 

[Gib Z]

Could you explain how you got that result? And the reason you don't know how to derive the indefinite integral is because it has no anti derivative, elementary or not (Other than the obvious define a function such that its derivative is tanx/x).

 

[zoki85]

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

Rereading what you repeatedly stated in your posts and what I said in mine, let me paraphrase you:Is this comment a joke?

I've just realized that I have been using the phrase 'improper integral' wrong - that's an integral that is convergent but unbounded. What I meant was an integral that has a 'principal value' - as Halls defined it but is not convergent. Sorry if this caused any confusion. As Dr. Edgar's recipe says, the sense in which a number can be applyed to this integral is exactly in the principal value sense.

Sometimes what makes intuitive sense isn't necessarily what is consistent with mathematical definition and concepts.In calculus,we must be extra careful what really means to calculate improper integrals and when we refer to convergence and indeterminate forms.The problem was ill-defined and that integral diverges.Period.What OP really wants,I suppose,is to compute infinite sum of corresponding 'Cauchy principal values' symmetrically about singular points. At each x=(2k+1)\pi /2 point (k=0,1,2,...) we have singularity.I think some people are wondering why improper integrals aren't defined to be Cauchy principal value?To repeat :the reason is they are not equivalent at all.The most important difference is improper integrals must satisfy more strict criterion for convergence then Cauchy's value,as required under the DEFINITION of integral as a Riemann sum.Not every function has a Cauchy principal value,and those that do,don't necessarily have a convergent improper integral.In the given example function f(x)=1/x increases/decreases without bound as x tends to 0 from both sides ,and talking about the value of integral over any interval (-a,a) is like saying \infty - \infty =0,which isn't necessarily the case as a discussion of indeterminate forms and l'hopital's rule reveal.Mere possibility of the graph symmetry becomes unimportant becouse of the fact we are dealing with indeterminate forms.Another way to look at it:What if you shifted the point real line is partitioned?Surely ,you wouldn't the value of improper integral to depend on where you choose this point.While it may be the case there is naturally obvious choice for functions like 1/x ,what about some polynomials of third degree which aren't neceassirily symmetric about point of origin ,or our function f(x)=(tan x)/x without symmetry with respect to any point x>0?Loosing the symmetry we are facing another serious problem regarding independence from underlaying coordinate system.Therefore,if we want to maintain a strict interpretation of the improper integral so that when exists,it is independent of the choice of the limit of integration, we must take care to destinguish Cauchy's principal value and convegence of improper integral.Violation of this requirement is inconsistent with definitions and developments of calculus,more so than trying to preserve (possible) aspects of the symmetry in some examples.It may be a good idea that professor Halls chime in again and reformulate OP's problem in a strict and meaningful math notation.

Regards,
Zoki

 

[Gib Z]

I Think the passage you quoted Dick on makes it clear that he knows and agrees what you stated after that quote, So not really required.

However I am sure that Dick knew in his mind that we were talking about the Principle Value, even if you think otherwise. When talking about an integral like this, it is just Obvious what DexterCioby wanted, the Principal Value, because a) If Dexter didn't want the principal value, he would already have known the problem, he is smart enough. b) He gave the solution, pi/2, which implies to everyone that he is talking about the Principal Value.

 

[zoki85]

When talking about an integral like this, it is just Obvious what DexterCioby wanted, the Principal Value, because a) If Dexter didn't want the principal value, he would already have known the problem, he is smart enough. b) He gave the solution, pi/2, which implies to everyone that he is talking about the Principal Value.

You don't read or follow the argument again.
It's not about singular (Principal value),but PLURAL:Infinite sum of "Cauchy Principal values".
How would I know that Dexter is smart enough,when he confused apples and oranges like that?

 

[Gib Z]

You don't read or follow the argument again.
It's not about singular (Principal value),but PLURAL:Infinite sum of "Cauchy Principal values".

Yes, my bad. I meant the Plural.

You have not been on the forums a long time. You will see over time that dexter is very knowledgeable. In fact When I saw that dextercioby had a problem with an integral, which I had never seen before, I knew that no one on these forums would be able to do it other than the exception of some of the mods, matt grime or mathwonk.

dexter always comes though with ingenious solutions for problems the rest of us are stumped with. This is for your information zoki85, dexter is smart enough.

 

[Dick]

Ok, so if we're all clear on the fact the integral has to be evaluated in the principal value sense, has anyone looked at Dr. Edgar's solution? It's not even that hard.

 

[siddharth]

Ok, so if we're all clear on the fact the integral has to be evaluated in the principal value sense, has anyone looked at Dr. Edgar's solution? It's not even that hard.

If I understood what Dr. Edgar said correctly, the residue at the simple poles comes out as -1/z_0 where z_0=(2*k+1)Pi/2 and thus cancels in pairs.

I didn't understand how tan(x+i*R) goes uniformly to i as R -> infinity.

 

[Curious3141]

If I understood what Dr. Edgar said correctly, the residue at the simple poles comes out as -1/z_0 where z_0=(2*k+1)Pi/2 and thus cancels in pairs.

I didn't understand how tan(x+i*R) goes uniformly to i as R -> infinity.

tan(x+iR) = (tan x + itanhR)/(1 - itanxtanhR)

limit of tanhR as R-> infty is 1. Simplify and the expression becomes i.

 

[zoki85]

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/

OK,but I don't see why is this formulation equivalent with our problem?

 

[Dick]

OK,but I don't see why is this formulation equivalent with our problem?

it's a contour integral? The limiting integral along the base of the rectangle as the rectangle is grown to infinity is the PV integral of tan(x)/x along the real axis.

 

[f(x)]

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.

Mathematica says its divergent.. :(

 

[Gib Z]

Read through the whole thread..you will see it...

 

[f(x)]

Read through the whole thread..you will see it...

LOL, i thought Mathematica > Maple

 

[Gib Z]

Well It many ways saying its divergent is actually much more professional than automatically finding the Principal Value. I'm sure if you could enter it to find the principal value mathematicia would be able to give it to you.