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

[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.