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

[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.