2

How can I evaluate the closed form of the following integral: $$\int_0^1\frac{x}{\ln(x+1)(x^3+3x+3)}dx$$

According to Wolfram Alpha, the numerical value of this integral is close to 0.2673, but it doesn't show up any closed form.

enter image description here

Mithlesh Upadhyay
  • 4,787
The Integral
  • 109
  • Do you know if the integral is supposed to have a nice closed form? – b00n heT May 14 '16 at 09:27
  • 1
    Anything is possible. Also I am expecting a closed form including some special functions. – The Integral May 14 '16 at 09:40
  • @TheIntegral, did you just come up with this integral at random? – Yuriy S May 14 '16 at 09:41
  • 1
    @YuriyS Would you consider this as a random integral? link – The Integral May 14 '16 at 09:46
  • 1
    @TheIntegral, what does 'random integral' even mean, and how is this related? I asked specifically if this integral is something you came up with, and what reasons do you have to expect a closed form? The polynomial in the denominator doesn't even have nice roots – Yuriy S May 14 '16 at 10:00

1 Answers1

0

I am almost sure that this integral does not have closed form solutions in terms of special functions. As a matter of fact even Feynmann would not be able to solve it as somebody else in here rightfully noted.

Yet if we were to modify the polynomial in the denominator so that it has nice roots then indeed we can come up with something. Consider the following integral instead:

\begin{eqnarray} J:=\int\limits_0^1 \frac{x}{\log(1+x) (x^3+3 x^2+3 x+1)} dx =? \end{eqnarray} Now define a following function: \begin{equation} I(b) := \int\limits_0^1 \frac{x (1+x)^b}{\log(1+x) (x^3+3 x^2+3 x+1)} dx \end{equation} subject to $I(-\infty)=0$. Then clearly $\left.I(b)\right|_{b=0} = J$. All we have to do now is to compute $I^{'}(b)$ and then integrate the result from minus infinity to zero. It turns out that the former can be done always, i.e. no matter what polynomial we have in the denominator whereas the later can be only done in very particular cases (meaning if the polynomial in question has nice roots). As usual by " can only be done in particular cases" I mean that I won't know how to do it given my scarce mathematical knowledge.

Let us proceed: \begin{eqnarray} I^{'}(b) &=& \int\limits_0^1 \frac{x (1+x)^b}{ (x^3+3 x^2+3 x+1)} dx\\ &=& \int\limits_0^1 x (1+x)^{b-3} dx\\ &\underbrace{=}_{u:=1/(1+x)}& \int\limits_{1/2}^1 \frac{1-u}{u^b} d u\\ &=& \frac{4-3 \cdot 2^b+2^b b}{4(b-1)(b-2)} \end{eqnarray}

Therefore the result reads: \begin{eqnarray} J = I(0) &=& \int\limits_{-\infty}^0 \frac{4-3 \cdot 2^b+2^b b}{4(b-1)(b-2)} db\\ &=& \log(2) + \frac{1}{4} \int\limits_0^\infty \exp(-\log(2) b)\left( \frac{1}{b+2} - \frac{2}{b+1}\right) db\\ &=& \log(2) - Ei(-2\log(2)) + Ei(-\log(2)) \end{eqnarray} where $Ei()$ is the exponential integral function.

Now if we try to do the same procedure for the other integral in the question above we will end up with integrating hypergeometric function with respect to their parameters which is something they definitely have not taught me in my primary school so I wouldn't be able to get any result.

Przemo
  • 11,331