6
$\begingroup$

I would like to solve the integral

$$\int\left(1-x^{p}\right)^{\frac{n-1}{p}}\log\left(1-x^{p}\right)dx.$$

My problem is that I arrived at a solution via wolfram alpha, but I would like to understand how one would arrive there by hand.

What I did is to use the substitution $x=\left(1-\exp(z)\right)^{\frac{1}{p}}$. This yields $$-\frac{1}{p}\int\left(1-\exp(z)\right)^{\frac{1}{p}-1}\exp\left(\frac{n-1+p}{p}z\right)zdz$$ for which wolfram alpha yields the integral

\begin{align} &\frac{\exp\left(z\frac{(n+p-1)}{p}\right)}{\left(n+p-1\right)^{2}}p\,_{3}F_{2}\left(1-\frac{1}{p},\frac{n}{p}-\frac{1}{p}+1,\frac{n}{p}-\frac{1}{p}+1;\frac{n}{p}-\frac{1}{p}+2,\frac{n}{p}-\frac{1}{p}+2;\exp(z)\right)\\&-z\frac{\exp\left(z\frac{(n+p-1)}{p}\right)}{\left(n+p-1\right)}\,_{2}F_{1}\left(\frac{p-1}{p},\frac{n+p-1}{p};\frac{n+2p-1}{p};\exp z\right). \end{align}

I would really like to understand how it arrived at the generalized hypergeometric function in the solution because I think it would be useful to spot it in an integral (like spotting possible solutions in terms of a gamma function or beta function in integrals).

Does anybody have a hint how it arrived at the solution?

$\endgroup$
7
$\begingroup$

You made a substitution $u=1-x^p$, transforming the integral as $$ \int \left(1-x^p\right)^{(n-1)/p} \log\left(1-x^p\right) \mathrm{d} x = - \frac{1}{p} \int \left(1-u\right)^a u^b \log\left(u\right) \mathrm{d} u $$ where $a = \frac{1}{p}-1$ and $b=\frac{n-1}{p}$. In order to integrate this we shall take advantage of the following differentiation property of the Gauss hypergeometric function: $$ \left( \frac{1}{\alpha_1} z \frac{\mathrm{d}}{\mathrm{d} z} + 1 \right) {}_2 F_1\left( \left. \begin{array}{cc} \alpha_1 & \alpha_2 \cr \beta&- \end{array} \right| z \right) = {}_2 F_1\left( \left. \begin{array}{cc} \alpha_1 +1 & \alpha_2 \cr \beta& - \end{array} \right| z \right) $$ Choosing $\beta=\alpha_1 + 1$ we get $$ \left( \frac{1}{\alpha_1} z \frac{\mathrm{d}}{\mathrm{d} z} + 1 \right) {}_2 F_1\left( \left. \begin{array}{cc} \alpha_1 & \alpha_2 \cr &\alpha_1+1& \end{array} \right| z \right) = {}_1 F_0\left( \left. \begin{array}{c} \alpha_2 \\ - \end{array} \right| z \right) = \left(1-z\right)^{-\alpha_2} \tag{1} $$ Now, using $$\left( \frac{1}{\alpha_1} z \frac{\mathrm{d}}{\mathrm{d} z} + 1 \right) f(z) = \frac{1}{\alpha_1} z^{1-\alpha_1} \frac{\mathrm{d}}{\mathrm{d} z} \left( z^{\alpha_1} f(z) \right) \tag{2} $$ Combining eqs. $(1)$ and $(2)$ and choosing $\alpha_2 = -a$ and $\alpha_1 = b +1$ we arrive at $$ \int \left(1-u\right)^a u^b \mathrm{d}u = \frac{u^{b+1}}{b+1} {}_2 F_1\left( \left. \begin{array}{cc} -a & b+1 \cr &b+2& \end{array} \right| u \right) \tag{3} $$ We can now integrate by parts: $$ \begin{eqnarray} \int \left(1-u\right)^a u^b \log(u) \mathrm{d}u &=& \frac{u^{b+1}}{b+1} \log\left(u \right) \cdot {}_2 F_1\left( \left. \begin{array}{cc} -a & b+1 \cr &b+2& \end{array} \right| u \right) \\ && - \frac{1}{b+1} \int u^b \cdot {}_2 F_1\left( \left. \begin{array}{cc} -a & b+1 \cr &b+2& \end{array} \right| u \right) \mathrm{d}u \end{eqnarray} $$ The latter integral is integrated in a similar way by using differentiation properties of generalized hypergeometric function: $$ \frac{\mathrm{d}}{\mathrm{d} z} \left( z^{\alpha_1} \cdot {}_3 F_2\left( \left. \begin{array}{ccc} \alpha_1 & \alpha_2 & \alpha_3 \cr \beta_1& \beta_2 & \end{array} \right| z \right) \right) = \alpha_1 z^{\alpha_1-1} \cdot {}_3 F_2\left( \left. \begin{array}{ccc} \alpha_1+1 & \alpha_2 & \alpha_3 \cr \beta_1& \beta_2 & \end{array} \right| z \right) $$ we get $$\begin{eqnarray} \int \left(1-u\right)^a u^b \log(u) \, \mathrm{d} u &=& \frac{u^{b+1}}{b+1} \log\left(u \right) \cdot {}_2 F_1\left( \left. \begin{array}{cc} -a & b+1 \cr &b+2& \end{array} \right| u \right) \\ && - \frac{u^{b+1}}{(b+1)^2} \cdot {}_3 F_2\left( \left. \begin{array}{ccc} -a & b+1 & b+1 \cr b+2& b+2 & \end{array} \right| u \right) \end{eqnarray} \tag{4} $$

$\endgroup$
2
  • $\begingroup$ Thanks for the answer. I have to run, but I'll go through it and accept once I understood everything. $\endgroup$ – fabee Jan 10 '13 at 17:13
  • $\begingroup$ Thanks again for the awesome answer. I would vote it up two if I could. $\endgroup$ – fabee Jan 11 '13 at 12:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.