I got a question concerning an integral. I need to know the analytical expression. I have to Mellin transforn a function and the integral is then sth. like this:

$$ \int x^{N-1} \frac{Ln(a -x)}{1-x} dx $$

where x $\in [0,1] $ and N is the Mellin moment and $a> 1$

Actually I have to integrate this from 0 to 1 regularizing the divergence with a "plus distribution" (known in particle physics), which means the full Integral will be sth. like

$$ \int_0^1 \left( x^{N-1} \frac{Ln(a -x)}{1-x} - \frac{Ln(a -1)}{1-x} \right)dx $$

So the divergent point is going to be subtracted. For me it would be fine somehow to get the result from the integral above. But Mathematica doesn't want to integrate it.

What I actually need is the Integral from 0 to 1 and then an Expansion in $N\rightarrow \infty $ and finally only collecting the finite terms independent of N.

Maybe someone knows a trick how to integrate those "beasts".

Sorry, did a mistake in the beginning. Now the integral is correctly regularized and shouldn't diverge.

And this is what I did:

Integrate[(x^(-1 + N) Log[a - x])/(1 - x) - Log[a - 1 ]/(1 - x), {x, 0, 1}, Assumptions -> a > 1]

Series[Normal[%], {N, Infinity, 0}]

Mathematica seems to do nothing.

Ah, and N can in general be complex. Here a Ref. for the Mellin transform: Mellin transform

  • $\begingroup$ Although a well written question you should really also post code of what you have tried and the error message (if any) that results. I have a question about the parameters a, b` and c. Do they belong to the set of real numbers? Are they positive or is zero and negative allowed as well. Is it fair to say that N is an integer? $\endgroup$ Commented Nov 16, 2015 at 0:18
  • $\begingroup$ I tried Integrate[ x^N (Log[a + b (x - c)] - Log[a - b c])/(1 - x), {x, 0, 1}, Assumptions -> b > 0 && c > 0 && a > b c && N \[Element] Integers && N >= 1] and got an error message that it did not converge. $\endgroup$ Commented Nov 16, 2015 at 0:36
  • $\begingroup$ Hi Jack,first of all thanks for your answer. I did a mistake writing the integral. Now it should converge, at least for x=1 the Integrand should give 0. And I removed some constants which are not necessary for me. $\endgroup$
    – Marcel
    Commented Nov 16, 2015 at 6:40
  • $\begingroup$ hm, nobody has any ideas? :( $\endgroup$
    – Marcel
    Commented Nov 18, 2015 at 9:40
  • $\begingroup$ This seems like the kind of question you'd be more likely to get an answer to at Mathematics, since I suspect it'd be more readily solved by some sort of contour integration technique or something, not by a clever way of using Mathematica. $\endgroup$
    – David Z
    Commented Aug 4, 2016 at 20:13

1 Answer 1


Denote the result with I. This logarithm has simple derivative which is 1/(a-x) with respect to a. then factor out the parameter a to get an integrant:

f(x) = x^(N-1) * (1-x)^ ((N+1)-N-1) / (1-(1/a)x) / (1-x)

If you integrate it from zero to one with respect to x you get a Lauricella FD function which is the same as an Appell F1 function, you choose which you like, Times a beta constant of B(N,1). If F(x) is the name of the integrated f(x)

F(a)=a*dI/da=B(N,1)*FD(N; 1,1;N+1 ;1,1/a) 

with a=N, b1=b2=1, c=N+1 and one argument is unity, the other 1/a

This is a function of the inverse a. Then you have to use an Euler transform to calculate a higher hypergeometric function of additional arguments + a constant. If the Euler transform is not convenient you may expand Lauricella FD to a formal series and formally integrate (again a constant needed)

Finally you have to check one point, I would suggest a=1 to expand directly the logarithm and calculate the integration constant. The singularity will show up when expanding the series and integrate the inverse a power (this does not evaluate at zero). Another way is to expand 1/(1-x) = Σ x^l1 and 1/(1-x/a) = Σ (x/a)^l2 Then the series is evaluated as:

F(a)=Σ (1/a)^(l2)/(N+l1+l2) 

Then integrate the general term of the series to get:

I = c-Σ (1/a)^(l2)/(N+l1+l2)/l2

Apart of the constant there is a logarithmic term at l2=0 so it correctly reads:

I = c-ln(a)* Σ 1/(N+l1) - Σ (1/a)^(l2+1)/(N+1+l1+l2)/(l2+1)

then it is formally right! I do not know the convergence conditions though. To calculate c set a=1 and expand ln(1+(1-x)) in powers of (1-x) Integrating term by term formally you get a beta function series. That can be used to find c.

Very nicely hidden intention with logarithms

  • 2
    $\begingroup$ How is that a Mathematica based answer? Can you back up your calculations with computations in Mathematica? $\endgroup$
    – corey979
    Commented Nov 30, 2020 at 23:55

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