Message Boards

WOLFRAM COMMUNITY

6549 Views

16 Replies

34 Total Likes

View groups...

Follow this post

Share this post:

GROUPS:

[CALL] for challenging univariate symbolic integrals

Oleg Marichev

Oleg Marichev, Wolfram Research Inc

Posted 1 year ago

This is a call for univariate indefinite or definite exact integrals that cannot immediately computed in Mathematica and Wolfram Language. Submitted integrals will be collected for potential new event where they will be discussed and solved. You may refer to our previous event, which is showcased on the Wolfram Community post: https://community.wolfram.com/groups/-/m/t/2821053. Do you have an integral that cannot be computed directly using Integrate ? Does the following apply to this integral? Submission rules DOs Univariate (terms in a single variable) Exact (use only symbols, integers, or exact fractions). Definite integrals (1, 2) Indefinite Integrals (1, 2) Keep comments directly related to the topics. Submission rules DON'Ts Don't submit those that Mathematica / Wolfram Language can do directly using Integrate[ ]. Don't submit numerical integrals (NIntegrate[ ]) Don't submit decimal approximate numbers in Integrate[ ]. Don't submit images, only copyable code. If you have such an integral - please put it in a comment below! Examples of submission formats You can add a notebook with your integral: You can post a formatted code: Integrate[Gamma[s] z^-s, s]

POSTED BY: Oleg Marichev

16 Replies

Sort By:

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 1 year ago

I tried another very popular book with random integrals with Mathematica that can't solve: Attachments: Define Integrals.nb

POSTED BY: Mariusz Iwaniuk

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 1 year ago

I opened another book about LaplaceTransform and tried a few examples: LaplaceTransform[BesselJ[v, a x] BesselI[v, b x], x, s, Assumptions -> {Re[v] > 0, a > 0, b > 0}](Can't Solve !) LaplaceTransform[(1 - Exp[-x])^v* Erfc[ a (1 - Exp[-x])], x, s, Assumptions -> {Re[v] > -1, a > 0}](Can't Solve !) LaplaceTransform[(1 - Exp[-x])^\[Mu]* Gamma[v, a(1 - Exp[-x])], x, s, Assumptions -> {Re[\[Mu]] > -1, Re[v] > 0, a > 0}](Can't Solve !) LaplaceTransform[(x + a)^nLegendreP[n, (x - a)/(x + a)], x, s, Assumptions -> {Re[n] > 0, a > 0}](Can't Solve !)

I opened another book about LaplaceTransform and tried a few examples:

      LaplaceTransform[BesselJ[v, a x] BesselI[v, b x], x, s, 
       Assumptions -> {Re[v] > 0, a > 0, b > 0}](*Can't Solve !*)

      LaplaceTransform[(1 - Exp[-x])^v* Erfc[ a (1 - Exp[-x])], x, s, 
       Assumptions -> {Re[v] > -1, a > 0}](*Can't Solve !*)

      LaplaceTransform[(1 - Exp[-x])^\[Mu]* Gamma[v, a*(1 - Exp[-x])], x, s,
        Assumptions -> {Re[\[Mu]] > -1, Re[v] > 0, a > 0}](*Can't Solve !*)

      LaplaceTransform[(x + a)^n*LegendreP[n, (x - a)/(x + a)], x, s, 
        Assumptions -> {Re[n] > 0, a > 0}](*Can't Solve !*)

POSTED BY: Mariusz Iwaniuk

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 1 year ago

I opened a book about MellinTransform and tried a few examples: MellinTransform[(Sqrt[x^2 + a^2] + a)^\[Rho]/Sqrt[x^2 + a^2]* BesselJ[v, bx], x, s, Method -> "Integrate", Assumptions -> {b > 0, Re[a] > 0, -Re[v] < Re[s] < 5/2 - Re[\[Rho]]}](Can't Solve !) MellinTransform[ Exp[-\[Sigma] Sqrt[x] - \[Omega]x] Hypergeometric1F1[a, b, \[Omega]x], x, s, Method -> "Integrate", Assumptions -> {a > 0, b > 0, Re[\[Sigma]] > 0, Re[s] > 0}](Can't Solve !) MellinTransform[ 1/(x + \[Sigma])^\[Rho] Hypergeometric2F1[a, b, c, -\[Omega]x], x, s, Method -> "Integrate", Assumptions -> {Re[\[Rho]] > 0, Re[\[Sigma]] > 0, Re[s] > 0}](Can't Solve !)

I opened a book about MellinTransform and tried a few examples:

    MellinTransform[(Sqrt[x^2 + a^2] + a)^\[Rho]/Sqrt[x^2 + a^2]*
     BesselJ[v, b*x], x, s, Method -> "Integrate", 
    Assumptions -> {b > 0, 
      Re[a] > 0, -Re[v] < Re[s] < 5/2 - Re[\[Rho]]}](*Can't Solve !*)

   MellinTransform[
    Exp[-\[Sigma] Sqrt[x] - \[Omega]*x] Hypergeometric1F1[a, 
      b, \[Omega]*x], x, s, Method -> "Integrate", 
    Assumptions -> {a > 0, b > 0, Re[\[Sigma]] > 0, 
      Re[s] > 0}](*Can't Solve !*)

   MellinTransform[
    1/(x + \[Sigma])^\[Rho] Hypergeometric2F1[a, b, 
      c, -\[Omega]*x], x, s, Method -> "Integrate", 
    Assumptions -> {Re[\[Rho]] > 0, Re[\[Sigma]] > 0, 
      Re[s] > 0}](*Can't Solve !*)

POSTED BY: Mariusz Iwaniuk

Henrik Schachner

Henrik Schachner, Radiation Therapy Center, Weilheim, Germany

Posted 1 year ago

I would be very interested in seeing a solution for this definite parameter integral: Integrate[((1 - x)^r (-2 x + (-2 + x) Log[1 - x]))/(2 x Log[1 - x]^2), {x, 0, 1}, Assumptions -> r >= 0]

POSTED BY: Henrik Schachner

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 1 year ago

Looks like: $$\int_0^1 \frac{(1-x)^r (-2 x+(-2+x) \log (1-x))}{2 x \log ^2(1-x)} \, dx=1+r-r \log (1+r)-\frac{1}{2} \log (2 \pi (1+r))+\text{log$\Gamma $}(1+r)$$ Integrate[((1 - x)^r (-2 x + (-2 + x) Log[1 - x]))/(2 x )t (1 - x)^ t, {t, 0, Infinity}, {x, 0, 1}, Assumptions -> {r >= 0, t >= 0}] (1 + r - r Log[1 + r] - 1/2 Log[2 \[Pi] (1 + r)] + LogGamma[1 + r]*) Regards M.I

Looks like: $$\int_0^1 \frac{(1-x)^r (-2 x+(-2+x) \log (1-x))}{2 x \log ^2(1-x)} \, dx=1+r-r \log (1+r)-\frac{1}{2} \log (2 \pi (1+r))+\text{log$\Gamma $}(1+r)$$

 Integrate[((1 - x)^r (-2 x + (-2 + x) Log[1 - x]))/(2 x )*t (1 - x)^
   t, {t, 0, Infinity}, {x, 0, 1}, Assumptions -> {r >= 0, t >= 0}]

 (*1 + r - r Log[1 + r] - 1/2 Log[2 \[Pi] (1 + r)] + LogGamma[1 + r]*)

Regards M.I

POSTED BY: Mariusz Iwaniuk

Michal Kvasnicka

Posted 1 year ago

Nice trick ... :)

POSTED BY: Michal Kvasnicka

Michal Kvasnicka

Posted 1 year ago

It should be nice to generalize Integrate function to be able to solve this directly: Integrate[((1 - x)^r (-2 x + (-2 + x) Log[1 - x]))/(2 x Log[1 - x]^2), {x, 0, 1}, Assumptions -> r >= 0] using (in this specific case) the following identity: 1/Log[1 - x]^2 = Integrate[t (1 - x)^t, {t, 0, Infinity}] if Re[Log[1 - x]] < 0 like Integrate[((1 - x)^r (-2 x + (-2 + x) Log[1 - x]))/(2 x)* t (1 - x)^t, {t, 0, Infinity}, {x, 0, 1}, Assumptions -> r >= 0]

It should be nice to generalize Integrate function to be able to solve this directly:

Integrate[((1 - x)^r (-2 x + (-2 + x) Log[1 - x]))/(2 x Log[1 - x]^2), {x, 0, 1}, Assumptions -> r >= 0]

using (in this specific case) the following identity:

1/Log[1 - x]^2 = Integrate[t (1 - x)^t, {t, 0, Infinity}] if Re[Log[1 - x]] < 0

Integrate[((1 - x)^r (-2 x + (-2 + x) Log[1 - x]))/(2 x)*
  t (1 - x)^t, {t, 0, Infinity}, {x, 0, 1}, Assumptions -> r >= 0]

POSTED BY: Michal Kvasnicka

Henrik Schachner

Henrik Schachner, Radiation Therapy Center, Weilheim, Germany

Posted 1 year ago

What a nice trick, indeed, I am impressed! Many thanks, Mariusz!

POSTED BY: Henrik Schachner

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 1 year ago

I have a site of Table of Define integrals but you have to go through the whole list and see if Integrate can solve. Attachments: Define integrals.nb

POSTED BY: Mariusz Iwaniuk

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 1 year ago

Another list for define integrals with attached files. Attachments: Table Całek Ozna...nb Table Całek Ozna...nb Table Całek Ozna...nb

POSTED BY: Mariusz Iwaniuk

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 1 year ago

Another list for define integrals with attached files. Examples from here. Attachments: Tintegrate integrals.nb

POSTED BY: Mariusz Iwaniuk

Frank Iannarilli

Posted 1 year ago

POSTED BY: Frank Iannarilli

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 1 year ago

I have a site of Table of indefine integrals but you have to go through the whole list and see if Integrate can solve. Another site which can be very helpful. My list attached below Attachments: 111432973.nb

POSTED BY: Mariusz Iwaniuk

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 1 year ago

2 another indefine integrals: Integrate[Exp[-a Sqrt[x]]Exp[-aSqrt[ x - b]], x](* 1/2 b (E^(-a Sqrt[b] x) (1/x - (2 + 2 a Sqrt[b] x + a^2 b x^2)/(a^3 b^(3/2))) + a Sqrt[b] ExpIntegralEi[-a Sqrt[b] x])) Integrate[Gamma[a , x]^2/x^2, x]( -((2 E^-x x^(-1 + a) Gamma[a, x])/(-1 + a)) + Gamma[a, x]^2/(-1 + a) + Gamma[a, x]^2/((-1 + a) x) - (a Gamma[a, x]^2)/((-1 + a) x) + ( 2^(2 - 2 a) Gamma[-1 + 2 a, 2 x])/(-1 + a)*)

2 another indefine integrals:

  Integrate[Exp[-a Sqrt[x]]*Exp[-a*Sqrt[ x - b]], x](* 1/2 b (E^(-a Sqrt[b]
        x) (1/x - (2 + 2 a Sqrt[b] x + a^2 b x^2)/(a^3 b^(3/2))) + 
     a Sqrt[b] ExpIntegralEi[-a Sqrt[b] x])*)

  Integrate[Gamma[a , x]^2/x^2, x](* -((2 E^-x x^(-1 + a) Gamma[a, x])/(-1 + a)) + Gamma[a, x]^2/(-1 + a) +
    Gamma[a, x]^2/((-1 + a) x) - (a Gamma[a, x]^2)/((-1 + a) x) + (
   2^(2 - 2 a) Gamma[-1 + 2 a, 2 x])/(-1 + a)*)

POSTED BY: Mariusz Iwaniuk

Roland Franzius

Posted 1 year ago

Hi Oleg, since a long time I don't come to terms with all the orthogonality relations of polynomial families. Just after the invention of the internet I found your books at Oxfords Blackwells,, ordered the four volumes and paid about 1000 GBP :-( The next days I found Mathenatica 2.0 in Hamburg for 50€ :-). Here is my attempt with Jacobi polynomials (Gradshtey/Ryzhik 7.391) (Notebook attached) Regards Roland Attachments: Integral submission.nb