Prev: Laplace transform of periodic function
Next: London Computational Knowledge Summit
From: pimeja on 16 Apr 2010 05:52
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.

From: David Park on 17 Apr 2010 06:02
Well, I'm treading on the edge of my knowledge, but what about doing a
series expansion?

First, evaluate the indefinite integral and then evaluate the result at the
limits.

intg0[x_] = Integrate[Log[x] Sin[x]/x, x];
Limit[%, x -> \[Infinity]] - Limit[%, x -> 0]

Then do a series expansion of the integral that Mathematica returned.

series1 =
Series[intg0[x], {x, 0, 7}, Assumptions -> x > 0] // Normal // Expand

Then do a series expansion of the initial integrand and integrate term by
term.

Series[Log[x] Sin[x]/x, {x, 0, 7}] // Normal
series2 = Integrate[#, x] & /@ % // Expand

The two series match term for term (at least as far as we have gone) so they
must represent the same function.


David Park
djmpark(a)comcast.net
http://home.comcast.net/~djmpark/



From: pimeja [mailto:sed.nivo(a)gmail.com]

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.



From: Leonid Shifrin on 17 Apr 2010 06:02
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:

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 integral
along the relevant half of the imaginary axis for each exponent separately
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 computation
in Mathematica
does confirm it.

Hope this helps.

Regards,
Leonid

On Fri, Apr 16, 2010 at 1:52 PM, pimeja <sed.nivo(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.
>
>
From: sashap on 17 Apr 2010 06:03
On Apr 16, 4:52 am, 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?

In[8]:= Limit[D[Integrate[x^(s - 1)*Sin[x], {x, 0, Infinity},
Assumptions -> -1 < s < 1], s], s -> 0]

Out[8]= -((EulerGamma*Pi)/2)

>
> 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.


From: pimeja on 18 Apr 2010 05:58
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.


 |  Next  |  Last
Pages: 1 2
Prev: Laplace transform of periodic function
Next: London Computational Knowledge Summit