I am trying to get an expression for this integral

$$ \int_0^1 \mathrm{d}x \dfrac{x^6\log(1-x)} {\sqrt{\left( x^3 - \lambda(x-1)\right)^3}} $$

with $ \lambda \ll1$.

I know this integral has no solution in terms of elementary functions so I am trying to see the behaviour in $\lambda$ when $\lambda \rightarrow 0$. I can't make a Taylor expansion for the square root in the denominator, because when $x \approx 0$, $\lambda$ and $x$ become comparable, and the expansion doesn't converge.

What I have tried is to split the integral in two intervals $[0,\lambda]$ and $[\lambda,1]$ and then expand for $x/\lambda \ll 1$ for the first interval and for $\lambda/ x \ll 1$ for the second one. However it seems I don't obtain the good behaviour. In fact, I'm not sure of the validity of this.

Any hint about how to get the good behaviour in $\lambda$?

Thanks in advance!


I am not sure I understand. When $\lambda=0$ the numerator is $-x^7$ since $\log(1-x)\sim -x$ and the denominator only $x^{9/2}$ so the integral converges. Can't you just use Lebesgue dominated convergence theorem?

EDIT OK, I understand now. Call $$ g(\lambda)=\int_{0}^{1}\frac{x^{6}\log(1-x)}{(x^{3}-\lambda(x-1))^{3/2}}dx $$ and

$$ f(x,\lambda)=\frac{x^{6}\log(1-x)}{(x^{3}-\lambda(x-1))^{3/2}}. $$ Then $$ \frac{\partial f}{\partial\lambda}(x,\lambda)=\frac{3}{2}\frac{(x-1)x^{6}% \log(1-x)}{\left( x^{3}+\lambda(1-x)\right) ^{\frac{5}{2}}}% $$ and so since $0\leq x^{3}\leq x^{3}+\lambda(1-x)$, \begin{align*} \left\vert \frac{\partial f}{\partial\lambda}(x,\lambda)\right\vert & =\frac{3}{2}\frac{(1-x)x^{6}|\log(1-x)|}{\left( x^{3}+\lambda(1-x)\right) ^{\frac{5}{2}}}\leq\frac{3}{2}\frac{(1-x)x^{6}|\log(1-x)|}{x^{\frac{15}{2}}% }\\ & \sim\frac{3}{2}x^{7-^{\frac{15}{2}}}=\frac{3}{2}x^{-^{\frac{1}{2}}}% \end{align*} near $x=0$, which is integrable (near $x=1$ there is no problem since $t\log t\to 0$), where we used the fact that $\log(1-x)\sim-x$ near $x=0$. So you can differentiate under the integral sign to find that $$ g^{\prime}(0)=\int_{0}^{1}\frac{3}{2}\frac{(x-1)x^{6}\log(1-x)}{\left( x^{3}\right) ^{\frac{5}{2}}}dx $$ and so$$ g(\lambda)=\int_{0}^{1}\frac{x^{6}\log(1-x)}{(x^{3})^{3/2}}dx+\lambda\int% _{0}^{1}\frac{3}{2}\frac{(x-1)x^{6}\log(1-x)}{\left( x^{3}\right) ^{\frac {5}{2}}}dx+o(\lambda) $$

  • $\begingroup$ Similarly the derivative of the integral with respect to $\lambda$ converges at $\lambda = 0$. You run into problems in trying to get higher orders than that. $\endgroup$ – Robert Israel Jul 12 '17 at 22:48
  • $\begingroup$ $\lambda$ is not zero, but small. At the end, I would like to have a function of $\lambda$. The difficult part, I think, is that here I can't just apply Taylor's formula for $\lambda$ small and then integrate, since for $x$ small there is an important contribution to the integral $\endgroup$ – ASM Jul 13 '17 at 11:27

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