Message Boards

WOLFRAM COMMUNITY

3460 Views

4 Replies

2 Total Likes

View groups...

Follow this post

Share this post:

GROUPS:

Mathematics Calculus Wolfram Language

How to solve this integral?

Masood Ahmed

Posted 4 years ago

Dear All, Could you please let me know how to solve this integration. Your support would be greatly appreciated. Integrate[ExpIntegralEi[-s] * Exp[(-s * w^2) / (1 + 2 * s * g^2)] / Sqrt[1 + 2 * s * g^2], {s,0, Infinity}] Attachments: Integration.png

POSTED BY: Masood Ahmed

4 Replies

Sort By:

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 4 years ago

This type integrals can be solved only by numerics: w = 1; g = 1; NIntegrate[ExpIntegralEi[-s]Exp[(-sw^2)/(1 + 2sg^2)]/Sqrt[1 + 2sg^2], {s, 0, Infinity}] (-0.669167 ) No hope for a closed-form solution by existing special function.

POSTED BY: Mariusz Iwaniuk

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 4 years ago

Using:$\sum _{j=0}^{\infty } \frac{x^j}{j!}=\exp (x)$ we can find solution by infinite Series: func = Integrate[(-((s w^2)/(1 + 2 g^2 s)))^j/j!* ExpIntegralEi[-s]/ Sqrt[1 + 2 g^2 s] // PowerExpand, {s, 0, Infinity}, Assumptions -> {j >= 0, w > 0, g > 0}] // FullSimplify // Expand (((-1)^j 2^(1 - j) g^(-2 - 2 j) j Sqrt[\[Pi]] w^(2 j) Gamma[j])/( j! Gamma[1/2 + j]) + ((-1)^(1 + j) 2^-j EulerGamma g^(-2 - 2 j) j Sqrt[\[Pi]] w^(2 j) Gamma[j])/( j! Gamma[1/2 + j]) + ((-1)^(1 + j) 2^-j g^(-2 - 2 j) j Sqrt[\[Pi]] w^(2 j) Gamma[j] HarmonicNumber[j])/( j! Gamma[1/2 + j]) + ((-1)^(1 + j) 2^(1/2 - j) g^(-1 - 2 j) Sqrt[\[Pi]] w^(2 j) HypergeometricPFQ[{-(1/2), 1/2 + j}, {1/2, 1/2}, 1/(2 g^2)])/ j! - ((-(1/2))^j g^(-4 - 2 j) Sqrt[\[Pi]] w^(2 j) Gamma[2 + j] HypergeometricPFQ[{1, 1, 2 + j}, {2, 2, 5/2}, 1/( 2 g^2)])/( 3 j! Gamma[1/2 + j]) + ((-1)^(1 + j) 2^-j g^(-2 - 2 j) j Sqrt[\[Pi]] w^(2 j) Gamma[j] Log[2])/( j! Gamma[1/2 + j]) + ((-1)^j 2^(1 - j) g^(-2 - 2 j) j Sqrt[\[Pi]] w^( 2 j) Gamma[j] Log[g])/(j! Gamma[1/2 + j])) func2 = Sum[func[[1 ;; 3]], {j, 0, Infinity}] + Sum[func[[6 ;; 7]], {j, 0, Infinity}] + Sum[func[[4 ;; 5]], {j, 0, Infinity}] // Simplify (Sum[((-1)^(1 + j)2^(1/2 - j)g^(-1 - 2j)Sqrt[Pi]w^(2j)HypergeometricPFQ[{-1/2, 1/2 + j}, {1/2, 1/2}, 1/(2g^2)])/j! - ((-1/2)^jg^(-4 - 2j)Sqrt[Pi]w^(2j)Gamma[2 + j]HypergeometricPFQ[{1, 1, 2 + j}, {2, 2, 5/2}, 1/(2g^2)])/(3j!Gamma[1/2 + j]), {j, 0, Infinity}] - (((2E^(w^2/(2g^2))g - Sqrt[2Pi]wErfi[w/(Sqrt[2]g)])(-2 + EulerGamma + Log[2] - 2Log[g]))/E^(w^2/(2g^2)) + 2gDerivative[1, 0, 0][Hypergeometric1F1][1, 1/2, -1/2w^2/g^2])/(2g^3)) Then: F[g_, w_, M_] := Sum[(((-1)^(1 + j) 2^(1/2 - j) g^(-1 - 2 j) Sqrt[\[Pi]] w^(2 j) HypergeometricPFQ[{-(1/2), 1/2 + j}, {1/2, 1/2}, 1/(2 g^2)])/ j! - ((-(1/2))^j g^(-4 - 2 j) Sqrt[\[Pi]] w^(2 j) Gamma[2 + j] HypergeometricPFQ[{1, 1, 2 + j}, {2, 2, 5/2}, 1/( 2 g^2)])/(3 j! Gamma[1/2 + j])), {j, 0, M}] - ( E^(-(w^2/( 2 g^2))) (2 E^(w^2/(2 g^2)) g - Sqrt[2 \[Pi]] w Erfi[w/(Sqrt[2] g)]) (-2 + EulerGamma + Log[2] - 2 Log[g]) + 2 g Derivative[1, 0, 0][Hypergeometric1F1][1, 1/2, -1/2w^2/g^2])/(2 g^3) // N F[1, 1, 100](Where M is should be a infinity in this case M=100) (-0.669167 + 0. I*) Looks like answer is the same.

Using:$\sum _{j=0}^{\infty } \frac{x^j}{j!}=\exp (x)$

we can find solution by infinite Series:

 func = Integrate[(-((s w^2)/(1 + 2 g^2 s)))^j/j!* ExpIntegralEi[-s]/
    Sqrt[1 + 2 g^2 s] // PowerExpand, {s, 0, Infinity}, 
  Assumptions -> {j >= 0, w > 0, g > 0}] // FullSimplify // Expand

(*((-1)^j 2^(1 - j) g^(-2 - 2 j) j Sqrt[\[Pi]] w^(2 j) Gamma[j])/(
 j! Gamma[1/2 + j]) + ((-1)^(1 + j) 2^-j EulerGamma g^(-2 - 2 j)
   j Sqrt[\[Pi]] w^(2 j) Gamma[j])/(
 j! Gamma[1/2 + j]) + ((-1)^(1 + j) 2^-j g^(-2 - 2 j) j Sqrt[\[Pi]]
   w^(2 j) Gamma[j] HarmonicNumber[j])/(
 j! Gamma[1/2 + j]) + ((-1)^(1 + j) 2^(1/2 - j) g^(-1 - 2 j)
   Sqrt[\[Pi]] w^(2 j)
   HypergeometricPFQ[{-(1/2), 1/2 + j}, {1/2, 1/2}, 1/(2 g^2)])/
 j! - ((-(1/2))^j g^(-4 - 2 j) Sqrt[\[Pi]] w^(2 j)
   Gamma[2 + j] HypergeometricPFQ[{1, 1, 2 + j}, {2, 2, 5/2}, 1/(
   2 g^2)])/(
 3 j! Gamma[1/2 + j]) + ((-1)^(1 + j) 2^-j g^(-2 - 2 j) j Sqrt[\[Pi]]
   w^(2 j) Gamma[j] Log[2])/(
 j! Gamma[1/2 + j]) + ((-1)^j 2^(1 - j) g^(-2 - 2 j) j Sqrt[\[Pi]] w^(
  2 j) Gamma[j] Log[g])/(j! Gamma[1/2 + j])*)

func2 = Sum[func[[1 ;; 3]], {j, 0, Infinity}] + 
   Sum[func[[6 ;; 7]], {j, 0, Infinity}] + 
   Sum[func[[4 ;; 5]], {j, 0, Infinity}] // Simplify
(*Sum[((-1)^(1 + j)*2^(1/2 - j)*g^(-1 - 2*j)*Sqrt[Pi]*w^(2*j)*HypergeometricPFQ[{-1/2, 1/2 + j}, {1/2, 1/2}, 1/(2*g^2)])/j! - 
   ((-1/2)^j*g^(-4 - 2*j)*Sqrt[Pi]*w^(2*j)*Gamma[2 + j]*HypergeometricPFQ[{1, 1, 2 + j}, {2, 2, 5/2}, 1/(2*g^2)])/(3*j!*Gamma[1/2 + j]), {j, 0, Infinity}] - 
 (((2*E^(w^2/(2*g^2))*g - Sqrt[2*Pi]*w*Erfi[w/(Sqrt[2]*g)])*(-2 + EulerGamma + Log[2] - 2*Log[g]))/E^(w^2/(2*g^2)) + 
   2*g*Derivative[1, 0, 0][Hypergeometric1F1][1, 1/2, -1/2*w^2/g^2])/(2*g^3)*)

Then:

 F[g_, w_, M_] := 
  Sum[(((-1)^(1 + j) 2^(1/2 - j) g^(-1 - 2 j) Sqrt[\[Pi]] w^(2 j)
         HypergeometricPFQ[{-(1/2), 1/2 + j}, {1/2, 1/2}, 1/(2 g^2)])/
       j! - ((-(1/2))^j g^(-4 - 2 j) Sqrt[\[Pi]] w^(2 j)
         Gamma[2 + j] HypergeometricPFQ[{1, 1, 2 + j}, {2, 2, 5/2}, 1/(
         2 g^2)])/(3 j! Gamma[1/2 + j])), {j, 0, M}] - (
    E^(-(w^2/(
       2 g^2))) (2 E^(w^2/(2 g^2)) g - 
        Sqrt[2 \[Pi]] w Erfi[w/(Sqrt[2] g)]) (-2 + EulerGamma + 
        Log[2] - 2 Log[g]) + 
     2 g Derivative[1, 0, 0][Hypergeometric1F1][1, 
       1/2, -1/2*w^2/g^2])/(2 g^3) // N

F[1, 1, 100](*Where M is should be a infinity in this case M=100*)
(*-0.669167 + 0. I*)

Looks like answer is the same.

POSTED BY: Mariusz Iwaniuk

Masood Ahmed

Posted 4 years ago

Thank you very much for your reply.

POSTED BY: Masood Ahmed

Masood Ahmed

Posted 4 years ago

I greatly appreciate your support. Thank you very much.

POSTED BY: Masood Ahmed

Reply to this discussion

Reply Preview

Attachments

Remove Add a file to this post

Follow this discussion

or Discard

Group Abstract

Feedback