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From: Leonid Shifrin on 19 Apr 2010 05:59
Hi,

I can only suggest a way to compute it with Mathematica, checking all this
manually is probably a harder task. First, you can still use the same
representation for the Log as a first order in epsilon expansion. Then,
shift the integration variable, to get an integral

Integrate[Abs[t]^eps/(t - alpha)*Sin[t - alpha], {t, -Infinity, Infinity}].

where alpha == b-a

This one Mathematica can compute separately for positive and negative half -
line:

In[7]:= Integrate[t^eps/(t - alpha)*Sin[t - alpha], {t, 0, Infinity},
Assumptions -> {0 < eps < 1, alpha > 0}]

Out[7]= -Gamma[-1 +
eps] (alpha Cos[
alpha - (eps \[Pi])/2] HypergeometricPFQ[{1}, {1 - eps/2,
3/2 - eps/2}, -(alpha^2/4)] + (-1 +
eps) HypergeometricPFQ[{1}, {1/2 - eps/2, 1 - eps/2}, -(alpha^2/
4)] Sin[alpha - (eps \[Pi])/2])

In[19]:= Integrate[(-t)^eps/(t - alpha)*Sin[t - alpha], {t, -Infinity,
0}, Assumptions -> {0 < eps < 1, alpha > 0}]

Out[19]= Gamma[-1 +
eps] (alpha Cos[
alpha + (eps \[Pi])/2] HypergeometricPFQ[{1}, {1 - eps/2,
3/2 - eps/2}, -(alpha^2/4)] + (-1 +
eps) HypergeometricPFQ[{1}, {1/2 - eps/2, 1 - eps/2}, -(alpha^2/
4)] Sin[alpha + (eps \[Pi])/2])

Note that I used an assumption alpha>0. You can also do a computation with
an opposite assumption,
but this is not necessary since the result is obviously an even function of
alpha. We then combine the two parts to get the answer:

total = -Gamma[-1 +
eps] (alpha Cos[
alpha - (eps \[Pi])/2] HypergeometricPFQ[{1}, {1 - eps/2,
3/2 - eps/2}, -(alpha^2/4)] + (-1 +
eps) HypergeometricPFQ[{1}, {1/2 - eps/2, 1 - eps/2}, -(
alpha^2/4)] Sin[alpha - (eps \[Pi])/2]) +
Gamma[-1 +
eps] (alpha Cos[
alpha + (eps \[Pi])/2] HypergeometricPFQ[{1}, {1 - eps/2,
3/2 - eps/2}, -(alpha^2/4)] + (-1 +
eps) HypergeometricPFQ[{1}, {1/2 - eps/2, 1 - eps/2}, -(
alpha^2/4)] Sin[alpha + (eps \[Pi])/2]);

and expand in <eps> as before, taking the coefficient of the first - order
term in <eps>:

In[24]:= res = FullSimplify(a)SeriesCoefficient [Series[total,{eps,0,1}],1]

Out[24]= -(1/2) \[Pi] (-1+2 EulerGamma+Cos[2 alpha]+Cos[alpha]
((HypergeometricPFQ^({0},{0,1},0))[{1},{1/2,1},-(alpha^2/4)]+(HypergeometricPFQ^({0},{1,0},0))[{1},{1/2,1},-(alpha^2/4)])+alpha
Sin[alpha]
((HypergeometricPFQ^({0},{0,1},0))[{1},{1,3/2},-(alpha^2/4)]+(HypergeometricPFQ^({0},{1,0},0))[{1},{1,3/2},-(alpha^2/4)]))

So, this seems to be the answer you want. You can check (at least
numerically) that you get back your previous simpler case when you set
alpha->0. Checking this all manually is another story...

Hope this helps.

Regards,
Leonid




On Sun, Apr 18, 2010 at 1:58 PM, pimeja <sed.nivo(a)gmail.com> wrote:

> Leonid,
>
> On Apr 17, 1:02 pm, Leonid Shifrin <lsh...(a)gmail.com> wrote:
> > Hi,
> >
> > I don't have a complete rigorous solution, just a couple of observations.
> >
> > First, you can represent Log[x]/x as a coefficient of the linear term in
> > <eps> of expansion x^(eps-1) around eps = 0:
>
> Thanks, this something more advanced than integration by parts :)
> Only issue I suffer with this is that I hoped to extend the approach
> to more interesting case:
> Integrate[
> Log[Abs[x - a]]*Sin[x - b]/(x - b), {x, -Infinity, Infinity}]
> This integral appears in solving boundary problems in semi-infinite
> domain.
> Would you please suggest something?
>
> >
> > In[7]:= Series[x^(eps-1),{eps,0,1}]
> >
> > Out[7]= 1/x+(Log[x] eps)/x+O[eps]^2
> >
> > Now, this integral I did not check in all details manually, but it
> should=
> be
> > not very hard to get
> > (by writing Sin[x] = 1/(2I)(Exp[I x] - Exp[-Ix]) and performing an inte=
> gral
> > along the relevant half of the imaginary axis for each exponent
> separatel=
> y
> > you should be able to get it)
> >
> > In[8]:= Integrate[x^(eps -1)*Sin[x],{x,0,Infinity}, Assumptions->
> > {0<=eps<1}]
> >
> > Out[8]= Gamma[eps] Sin[(eps \[Pi])/2]
> >
> > Expanding this around <eps> = 0 and picking the first - order term, we =
> get
> > the result:
> >
> > In[10]:= Series[Gamma[eps] Sin[(eps \[Pi])/2],{eps,0,1}]
> >
> > Out[10]= \[Pi]/2-1/2 (EulerGamma \[Pi]) eps+O[eps]^2
> >
> > which is, by the way, twice smaller than you quoted - and direct
> computat=
> ion
> > in Mathematica
> > does confirm it.
> >
> > Hope this helps.
> >
> > Regards,
> > Leonid
> >
> >
> >
> > On Fri, Apr 16, 2010 at 1:52 PM, pimeja <sed.n...(a)gmail.com> wrote:
> > > Hi All,
> >
> > > For Integrate[Log[x] Sin[x]/x, {x, 0, \[Infinity]}] Mathematica
> > > returns -EulerGamma \[Pi].
> > > How to proof this analytical?
> >
> > > I've tried to use residue theory but it seems not suitable since
> > > integrand has pool of second order in zero (for Jordan lema should be
> > > first order only). Substitution x=Exp[y] returns even more strange
> > > result.
> >
> > > Thanks in advance.
>
>
>
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