Integrate a transcendental function given by a power series?

Questions
1

Is it possible to integrate a transcendental function given by a power series sum, instead of summing up the individual integrals in the power series? Reason why I’m asking this is cause each individual integral takes too long to calculate, so if I need to sum say 50 terms it will take forever, so if Mathematica has a way to define a function by means of a power series, however complicated it may get and then calculate its integral over say 0 and 1, that’d be ideal.

Let me give an example below:

The example is such that summing up 50 terms of this series would require calculating 50 integrals which are really slow to calculate, so I think there must be a better way to do that.

Let me know if you’d like to me to paste the plain Mathematica code in here, I’m new to this site so I don’t know very well how to do that.

M = 10; n = 2; m = 1; 
  N[(1/4)*
  Sum[(I*2*Pi*n)^k*Sum[(((-2)^(k-p)*(m*p)^(k+1))/(p!*(k-p)!*k!))* 
  Integrate[(1-u)^k*Coth[(m*p*u)/2]*(E^(m*p*n*u)-1), {u,0,1}], {p,1, k}], {k,1,M}], 5]

23145etwgdfs.JPG

Btw, how do I copy and paste here from a notebook without having to replace Mathematica’s graphic symbols with plain commands?

2

Cell, convert to,input form and copy.

 M = 10; n = 2; m = 1; 
  N[(1/4)*
  Sum[(I*2*Pi*n)^k*Sum[(((-2)^(k - p)*(m*p)^(k + 1))/(p!*(k - p)!*k!))*
              
      Integrate[(1 - u)^k*Coth[(m*p*u)/2]*(E^(m*p*n*u) - 1), {u, 0, 
        1}], {p, 1, k}], {k, 1, M}], 5]

Have you tried?

   M = 10; n = 2; m = 1; 
  (1/4)*Sum[(I*2*Pi*n)^k*
   Sum[(((-2)^(k - p)*(m*p)^(k + 1))/(p!*(k - p)!*k!))*
             
     NIntegrate[(1 - u)^k*Coth[(m*p*u)/2]*(E^(m*p*n*u) - 1), {u, 0, 
       1}, WorkingPrecision -> 20], {p, 1, k}], {k, 1, M}]

 (*-6.2667948461439727*10^7 + 1.89297162946885025*10^7 I*)
3

Not yet, I will try this, it should be just what I need. Thank you so much, you’re the best :slight_smile:

Edit: Yes, it runs faster.

Btw, for my learning, what in your code is different from the formula in the picture that makes it run faster?

4

NSum uses floating point arithmetic, so it doesn’t have to worry about the algebra!

I think the code can be sped up even more!

Mathematica is really very fast, if you use it right. See how much I sped up the MRB constant record computations !!!

5

Amazing, that is the best. If I’m able to finish my work successfully I will let you know so you can have a look (no more hassle for you, just for your curiosity).

6

Hey Marvin, the below is the actual sum that I need to calculate.

I see two problems with this approach.

  • It’s still not fast enough
  • it’s not giving me a very precise result as when I use say
    N[*,10], E.g., it displays the result as 0. * 10^11 (no decimal
    places)

Do you think you might be able to optimize this more and make the result more precise?

M = 100; m = Log[2]; N[
 x5 = Sum[(-1)^k*(2*Pi)^(2*k)*(2*k)!*
    Sum[(BernoulliB[2*j]/((2*k + 1 - 2*j)!*(2*j)!))*
              
      Sum[((-2)^(2*k - 
             p)/(p!*(2*k - p)!))*((-(1/
               2))*((m*p)^(2*j + 2)/(2*j + 2)!) - 
                     ((m*p)^(2*j + 1)/(2*j + 1)!)*Log[(m*p)/(2*Pi)] + 
          Sum[((m*p)^(2*j + 2 - i)*Zeta[i])/
                         (2*j + 2 - i)!, {i, 2, 
            2*j + 2}] - (1/2)*((m*p)^(2*j + 2)/(2*j + 1)!)*
                       
           Integrate[(1 - u)^(2*j + 1)*
              Coth[(m*p*u)/2] - ((2*Pi)/(m*p))*(1 - u)*Cot[Pi*u], {u, 
             0, 1}]), 
                {p, 1, 2*k}], {j, 0, k}], {k, 1, M}], 10]
7

Is this actually the formula you wan’t an answer too?

help1.JPG
help1.JPG1243×131 17 KB

Look at it carefully before we decide to work on it.

8

Let me check and I’ll get back to you. Hang on. Thanks a lot btw

9

Let me give you a little background first.
The formula I’m trying to calculate is below:

form1.JPG

Note it converges if M is finite, but not infinite. But then, below I have a formula which gives an analytic continuation for the infinite sum over n which converges even if m>0 (if m<0 both the left and the right-hand sides converge and they match, if m>0 only the right-hand side converges – note that below M=\infty):

form2.JPG
form2.JPG1047×131 17.2 KB

Now, to make things simpler, let’s try to sum up everything together for a large M2. Let’s ignore the complex part, because for positive m only the last term is complex. I could’ve used better variable names, but darn Mathematica made N a reserved keyword.

Hence altogether the sum I need to calculate is this little monster:
form3.JPG

And the input formula is:

M2 = 100; m = Log[2]; 
  N[Sum[(-1)^k*(2*Pi)^(2*k)*(2*k)!*
   Sum[(BernoulliB[2*j]/((2*k + 1 - 2*j)!*(2*j)!))*
            
     Sum[((-2)^(2*k - 
            p)/(p!*(2*k - p)!))*((-(1/
              2))*((m*p)^(2*j + 2)/(2*j + 2)!) - ((m*p)^(2*j + 
               1)/(2*j + 1)!)*Log[(m*p)/(2*Pi)] + 
                   
         Sum[((m*p)^(2*j + 2 - i)*Zeta[i])/(2*j + 2 - i)!, {i, 2, 
           2*j + 2}] - (1/2)*((m*p)^(2*j + 2)/(2*j + 1)!)*
                     
          Integrate[(1 - u)^(2*j + 1)*
             Coth[(m*p*u)/2] - ((2*Pi)/(m*p))*(1 - u)*Cot[Pi*u], {u, 
            0, 1}]), {p, 1, 2*k}], {j, 0, k}], {k, 1, M2}], 10]
10

I hate it when formulas get that messy.(Sometimes they have to, but by being clever, we can, sometimes, simplify them a little. There are even ways of making Mathematica simplify it.) Let me see if I can come up with a less complicated one.
Where did the formula that you said diverges at infinity come from?

11

I think you won’t be able to simplify that long formula further.
I even posted a post on math.stackexchange asking if somebody knows how to simplify it.
Besides, I know the Simplify formula in Mathematica, it just doesn’t work for these sums.

The link is below: Post on MSE