2

This is the original question: $$I=\int_{1}^{\sqrt{2}}\ln\left(\frac{2-2x^{2}+x^{4}+2\left(ex^{x}\right)^{\frac{1}{x}}-2x^{2}e^{\frac{1}{x}}+\left(ex^{3x}\right)^{\frac{1}{x}}}{2x-2x^{2}+x^{3}}\right)dx$$

This is essentially reducible to: $$I=\int_{1}^{\sqrt{2}}\ln\left(\frac{x^{4}-2x^{2}+2}{x\left(2-2x+x^{2}\right)}+e^{\frac{1}{x}}\right)dx$$

I thought of using Series Expansion of $\ln(1+x)$ as follows: $$I=\ln\left(\sqrt{2}\right)-\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k}}{k}\left(\int_{1}^{\sqrt{2}}\left(\frac{x^{4}-2x^{2}+2}{e^{\frac{1}{x}}x\left(2-2x+x^{2}\right)}\right)^{k}dx\right)$$ But I do not know how to continue from here.

The other method I tried was separating the numerator and denominator as follows:

$$\int_{1}^{\sqrt{2}}\ln\left(2-2x^{2}+x^{4}+2\left(ex^{x}\right)^{\frac{1}{x}}-2x^{2}e^{\frac{1}{x}}+\left(ex^{3x}\right)^{\frac{1}{x}}\right)dx-\int_{1}^{\sqrt{2}}\ln\left(2x-2x^{2}+x^{3}\right)dx$$

The integral on the right side can be evaluated and has the following answer (Which I got using CAS): $$\int_{1}^{\sqrt{2}}\ln\left(2x-2x^{2}+x^{3}\right)dx=\frac{\pi}{4}+\frac{\ln2}{\sqrt{2}}+\left(\sqrt{2}-1\right)\left(\ln\left(4-2\sqrt{2}\right)-3\right)$$

This is a question from a source and I have confirmed that the answer is in the form of $Li_2(x)$ and $\Gamma(x)$.

Miracle Invoker
  • 3,200
  • 1
  • 3
  • 22
  • Is there a reason you expect this integral to have a closed form answer? – Ninad Munshi Aug 03 '23 at 07:47
  • @NinadMunshi Well as stated in the end, it does have a closed form, but it is not elementary. – Miracle Invoker Aug 03 '23 at 07:48
  • @NinadMunshi How did you arrive at that conclusion? It is not matching numerically. – Miracle Invoker Aug 03 '23 at 07:49
  • You're right, I simplified incorrectly – Ninad Munshi Aug 03 '23 at 07:50
  • At the very least, $e^{1/x}$ won't help you unless you have some Feynman trick in mind. The integral is equivalent to $$\int_{\frac{1}{\sqrt{2}}}^1 \log\left(\frac{2x^4-2x^2+1}{x(2x^2-2x+1)}+e^x\right)\frac{dx}{x^2}$$ and there might be some symmetric significance to having a rational polynomial of the form $\frac{f(x^2)}{xf(x)}$ inside the log. – Ninad Munshi Aug 03 '23 at 07:55
  • It is sad that we do not know the general expression of the coefficients of the Taylor series of the integrand around $x=1$ because the convergence is quite fast. – Claude Leibovici Aug 03 '23 at 09:21
  • 1
    What is the source of this integral? The integral without the $e^{1/x}$ and with an $\frac{1}{x}$ appeared here on MSE before. – Zacky Aug 03 '23 at 10:44
  • @Zacky Well it's posted on a Math Server. And could you link that problem, maybe we could get some hint seeing its solution. – Miracle Invoker Aug 03 '23 at 11:24
  • @Zacky. I also have the feeling it saw the one you mention but I am unable to find it. Do you remember where ? – Claude Leibovici Aug 03 '23 at 11:29
  • @BlackEmperor see here: https://math.stackexchange.com/q/3005764/515527 - the $\frac1x$ is actually outside of the log. Can you also share your math forum? (maybe that can help) – Zacky Aug 03 '23 at 12:39
  • 2
    @BlackEmperor Can you please link the source giving the closed form with $\text{Li}_2(x)$ and $\Gamma(x)$? – Тyma Gaidash Aug 03 '23 at 17:00
  • Replace $\color{red}{\large +} e^{1/x}$ by $\color{red}{\large \cdot} e^{1/x}$ – Claude Leibovici Aug 04 '23 at 05:36
  • Numerically checking the “$+e^\frac1x$” version, it matches the original problem, so there is probably no “$\cdot e^\frac1x$”. – Тyma Gaidash Aug 04 '23 at 11:50

0 Answers0