Yes, especially the simplification given by Robert Israel (now having only
1 parameter in the integral): reading that 'delta' as 'moneyness' and the
integration variable as (normed) volatility that exp term is something like
the Vega of a normed Call.

Thus it may be that Alejandro's suggestion practical may work: usually the
moneyness is close to zero and volatility < 1 (which means, that the series
may converge for practical data, cutting off the integral).

But since it is for approximating ... approximating may be faster anyway
and sufficient.

radius of convergence = 1, no?

I think over R you would need hypergeom instead of a power series

radius of convergence = 1, no?

I think over R you would need hypergeom instead of a power series

So you want a symbolic solution ...I think one can reduce that to find

Int(t^2/(t^2+delta*mu)*exp(-1/4/mu*t^2-1/4*mu/t^2*delta^2),t = 0 .. infinity)

but then I gave up, may be Gradshteyn or similar can give you more

So you want a symbolic solution ...I think one can reduce that to find

Int(t^2/(t^2+delta*mu)*exp(-1/4/mu*t^2-1/4*mu/t^2*delta^2),t = 0 .. infinity)

but then I gave up, may be Gradshteyn or similar can give you more

Manoel,

More or less I understand what you want.

My question is: why do you think your integral can be expressed in terms of known function?
It is not unlikely that this is not possible (or extremely complicated).

I think you have to solve it numerical (which means: you need numerical values for mu and delta)
and as acer says you have to use Pi for 3.14159... with capital P not pi with lower case p.

For example:

  restart;
  Digits:=14;
  assume(0 < mu, 0 < delta);
  J:=Int((exp(1/sqrt(2/delta)^2)/((t+delta*mu/(delta+1))^2*(2*sqrt(2/delta)*
         sqrt(2*Pi*delta*mu/(delta+1)))*t^(3/2))*(t+delta*mu/(delta+1)))*
         exp(-(t*(delta+1)/(delta*mu)+delta*mu/((delta+1)*t))/(2*sqrt(2/delta)^2)), 
           t = 0 .. infinity);
  simplify(%);
 
  eval(J, [mu=1,delta=1]);
  evalf(%);
 
                           1.2421278438587

Manoel,

More or less I understand what you want.

My question is: why do you think your integral can be expressed in terms of known function?
It is not unlikely that this is not possible (or extremely complicated).

I think you have to solve it numerical (which means: you need numerical values for mu and delta)
and as acer says you have to use Pi for 3.14159... with capital P not pi with lower case p.

For example:

  restart;
  Digits:=14;
  assume(0 < mu, 0 < delta);
  J:=Int((exp(1/sqrt(2/delta)^2)/((t+delta*mu/(delta+1))^2*(2*sqrt(2/delta)*
         sqrt(2*Pi*delta*mu/(delta+1)))*t^(3/2))*(t+delta*mu/(delta+1)))*
         exp(-(t*(delta+1)/(delta*mu)+delta*mu/((delta+1)*t))/(2*sqrt(2/delta)^2)), 
           t = 0 .. infinity);
  simplify(%);
 
  eval(J, [mu=1,delta=1]);
  evalf(%);
 
                           1.2421278438587

I neither like the search nor saving a thread "as is" (it should kick off the boundaries)

Meanwhile mostly I save them by printing as pdf using PDFCreator.

If I really can overcome my paranoia for a local Google search that would do.

But anyway: the search here is ... ok, let me call it 'oldish'

not automatically, but for example this way(s)

  assume(0 < phi, phi < 1*Pi);
  SphericalY(1,1,phi,theta);
  convert(%,elementary);
  combine(%, radical);
  simplify(%, trig);

                                       1/2
                  -1/4 I exp(theta I) 6    sin(phi)
                  ---------------------------------
                                  1/2
                                Pi


  assume(0 < phi, phi < 2*Pi);
  SphericalY(1,1,phi,theta);
  convert(%,elementary);
  combine(%, radical);
  simplify(%, trig);
  simplify(%);

                                     1/2
                -1/4 I exp(theta I) 6    | sin(phi) |
                -------------------------------------
                                  1/2
                                Pi

not automatically, but for example this way(s)

  assume(0 < phi, phi < 1*Pi);
  SphericalY(1,1,phi,theta);
  convert(%,elementary);
  combine(%, radical);
  simplify(%, trig);

                                       1/2
                  -1/4 I exp(theta I) 6    sin(phi)
                  ---------------------------------
                                  1/2
                                Pi


  assume(0 < phi, phi < 2*Pi);
  SphericalY(1,1,phi,theta);
  convert(%,elementary);
  combine(%, radical);
  simplify(%, trig);
  simplify(%);

                                     1/2
                -1/4 I exp(theta I) 6    | sin(phi) |
                -------------------------------------
                                  1/2
                                Pi

how would you make your question more precise?

what would you tell a student how the answer should look like?

how would you make your question more precise?

what would you tell a student how the answer should look like?

such behaviour would make me to run security software and possibly to delete and re-install Firefox as well as considering noscript, noscript.net/. And I would only use the original from Mozilla, no advertising Firefox ...

The shortcut would be <ctrl> + t for a new tab (which should open with the deafault page I think)

otherwise provide complete replies (and Maple versions etc, best is a worksheet, classical *.mws format).

gn8

otherwise provide complete replies (and Maple versions etc, best is a worksheet, classical *.mws format).

gn8