a definite integral and a workaround

• To: mathgroup at smc.vnet.net
• Subject: [mg78380] a definite integral and a workaround
• From: dimitris <dimmechan at yahoo.com>
• Date: Fri, 29 Jun 2007 05:53:21 -0400 (EDT)

I have posted some time ago a similar message
but in view of the recent thread called "two integrals"
here we go...

In[1]:=
\$Version

Out[1]=
"5.2 for Microsoft Windows (June 20, 2005)"

I am interested in the integral

In[2]:=
Integrate[u*Sin[u*x]*Exp[(-y)*(Sqrt[1 + m^2*u^2]/m)], {u, 0,
Infinity}, GenerateConditions -> False]

Out[2]=
Integrate[(u*Sin[u*x])/E^((Sqrt[1 + m^2*u^2]*y)/m), {u, 0, Infinity},
GenerateConditions -> False]

I.e. it stays unevaluated within Mathematica (the same holds true for
another CAS I tried).

Say,

In[6]:=
integrand = u*Sin[u*x]*Exp[(-y)*(Sqrt[1 + m^2*u^2]/m)]

Out[6]=
(u*Sin[u*x])/E^((Sqrt[1 + m^2*u^2]*y)/m)

Then,

In[26]:=
ff=Integrate[integrand, x, y]
Normal[Series[%, {y, 0, 5}]]
(Integrate[#1, {u, 0, Infinity}, GenerateConditions -> False] & ) /@ %

Out[26]=
(m*Cos[u*x])/(E^((Sqrt[1 + m^2*u^2]*y)/m)*Sqrt[1 + m^2*u^2])

Out[27]=
(m*Cos[u*x])/Sqrt[1 + m^2*u^2] - y*Cos[u*x] + (Sqrt[1 +
m^2*u^2]*y^2*Cos[u*x])/(2*m) - ((1 + m^2*u^2)*y^3*Cos[u*x])/(6*m^2) +
((1 + m^2*u^2)^(3/2)*y^4*Cos[u*x])/(24*m^3) - ((1 +
m^2*u^2)^2*y^5*Cos[u*x])/(120*m^4)

Out[28]=
(m*BesselK[0, Sqrt[x^2]/Sqrt[m^2]])/Sqrt[m^2] - (y^2*BesselK[1,
Sqrt[x^2]/Sqrt[m^2]])/(2*m*Sqrt[x^2]) +
(Sqrt[m^2]*y^4*BesselK[2, Sqrt[x^2]/Sqrt[m^2]])/(8*m^3*x^2)

Pay notice to the last output. Next

In[31]:=
FullSimplify /@ Normal[Series[BesselK[0, Sqrt[x^2 + y^2]/m], {y, 0,
5}]]

Out[31]=
BesselK[0, Sqrt[x^2]/m] - (y^2*BesselK[1, Sqrt[x^2]/m])/
(2*m*Sqrt[x^2]) + (y^4*BesselK[2, Sqrt[x^2]/m])/(8*m^2*x^2)

Comparing the outputs 28 and 31 we conclude that

Integrate[ff,{u,0,Infinity}]=BesselK[0, Sqrt[x^2 + y^2]/m]

so the desiring integral is given by

In[33]:=
FunctionExpand[D[BesselK[0, Sqrt[x^2 + y^2]/m], x, y]]

Out[33]=
(x*y*BesselK[2, Sqrt[x^2 + y^2]/m])/(m^2*(x^2 + y^2))

Let's check it!

In[44]:=
NIntegrate[u*Sin[u*x]*Exp[(-y)*(Sqrt[1 + m^2*u^2]/m)] /. {x -> 1/2, m -
> 2, y -> 2}, {u, 0, Infinity}]

Out[44]=
0.08892633231521632

In[48]:=
(x*y*BesselK[2, Sqrt[x^2 + y^2]/m])/(m^2*(x^2 + y^2)) /. {x -> 1/2, m -
> 2, y -> 2}
N[%]

Out[48]=
(1/17)*BesselK[2, Sqrt[17]/4]
Out[49]=
0.08892633231440279

Dimitris

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