Matematické Fórum

Skrytý text:
a ten nekonverguje.

Skrytý text:

Let ; this improper integral is clearly convergent (both limits are finite at bounds).

First part: It is not hard to show, that .

Since , one can use substitution in the given integral, yielding the integral . Now it's enough to apply integration by parts.

Second part: Evaluate using conjugate Fourier series of a suitable function.

Let for , and extend this definition to so that is -periodic. Obviously, is odd therefore Fourier coefficients are all zero. For we have
.
So, by the definition, the conjugate Fourier series of is of the form:
, specially .
Now we use the fineteness of to compute the conjugate function of at (it would be much harder anywhere else anyway):
.
The equality then follows from the analogue of Dini's criterion for classical Fourier series (it' s enough to check, that is integrable over period and that is finite, what we did), so we get the result .


P.S. I'm currently writing my thesis on conjugate Fourier series so this was easy for me, though I know the solution might look quite extraordinary. I'm very interested in other methods you have to evaluate this for comparison. I will gladly provide more details on the criterion I used or on conjugate series, conjugate function in general if you like.

Skrytý text:

The equality is just definition of conjugate function to f at x (it is sufficient to work with -periodic functions only).

It arises naturally as a candidate for a function, to which the conjugate series might converge under certain assumptions for f. This can be seen easily from formula for partial sums of , where are Fourier coef. of f, which can be derived in a similar way as for the classical Fourier series (conjugate Dirichlet kernel is sum of sines):

One rather obvious way to ensure convergence of both integrals (after splitting by the minus between cosines above) is to presume convergence of - the first integral is then the conjugate function (the limit in def. now being unnecessary) and second integral tends (uniformly) to 0 by Riemann-Lebesgue lemma. This is exactly the Dini criterion I used. It is primitive but telling - one can see that the convergence of conj. series depends on local behaviour of f, whereas the sum depends on all values of f.

But this still doesn't justify the limit in the definition. Well, without it it is quite obvious that for a general integrable function, the definition would often fail. With the limit however, it can be seen from more advanced criteria that the convergence of conj. series and existence of conj. function are equivalent under certain (not so strong) assumptions for f as for example bounded variation of f or that the point is the Lebesgue point of f.

This leads us to question when the conjugate function (with the limit) exist itself. Well, it is known that it exists almost everywhere for any integrable f, though the proof of it is really difficult (for function from L2 is quite easy). However, it is remarkable result because it can't be achieved by theory of integration and isn't obvivous even for for example continous f.