$$I:=\int_0^\infty \ln x\,\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) \frac{x^4+3\alpha x^2- 1}{x^6}dx=\frac{(1+\alpha)\sqrt{2\alpha^3 \pi}}{2\sqrt[\alpha]e},\qquad \alpha>0.$$

This one looks very nice. It has stumped me.

Differentiation with respect to parameter does not seem to work either if I try $I(\alpha)$ and $I'(\alpha)$. at x=0 there seems to be a problem with the integrand also however I am not sure how to go about using this. Perhaps we could try and use a series expansion for $e^x=\sum_{n=0}^\infty x^n /n!$, however the function $e^{-1/x^2}$ is well known that its taylor series is zero despite the function not being.

  • $\begingroup$ Where on earth are you getting these from? $\endgroup$ May 3, 2014 at 6:13
  • $\begingroup$ I already asked you this question to which you never answered : are you planning to produce a textbook with "nice" integrals ? $\endgroup$ May 3, 2014 at 6:20
  • 4
    $\begingroup$ I fyou are old, what am I ? My adress is Jurassic.Park $\endgroup$ May 3, 2014 at 6:28
  • 1
    $\begingroup$ $[Energy,More]\neq 0$ @MhenniBenghorbal where [F,G]=FG-GF. $\endgroup$ May 3, 2014 at 22:04
  • 1
    $\begingroup$ I suggest substituting $y=x-1/x$. $\endgroup$
    – Chen Wang
    May 4, 2014 at 8:45

1 Answer 1


Assuming the $3\alpha x$ term is in fact $3\alpha x^2$ (otherwise the numerical results do not match).

$$\begin{align*} I&=\int_0^\infty \ln x\,\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) \frac{x^4+3\alpha x^2- 1}{x^6}dx\\ &=\int_0^\infty \ln x\, d\left(-\alpha x^{-3}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)\right)\\ &=-\alpha\left(\left.\frac{\ln x}{x^3}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)\right|_0^\infty-\int_0^\infty \frac{1}{x^3}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) d\,\ln x\right)\\ &=\alpha\int_0^\infty \frac{1}{x^4}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) dx\\ &=\alpha\left(\int_0^1 \frac{1}{x^4}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx+\underbrace{\int_1^\infty \frac{1}{x^4}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx}_{x\to1/x}\right) \\ &=\alpha\left(\int_0^1 \frac{1}{x^4}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx+\int_1^0 -x^2\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx\right) \\ &=\alpha\int_0^1 (x^2+x^{-4})\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx\\ &=\alpha\int_0^1 (x^2-1+x^{-2})\exp\left(-\frac{1}{\alpha}-\frac{(x-1/x)^2}{2\alpha }\right)d(x-1/x)\\ &=\alpha e^{-1/\alpha}\int_0^1 (1+(x-1/x)^2)\exp\left(-\frac{(x-1/x)^2}{2\alpha }\right)d(x-1/x)\\ &=\alpha e^{-1/\alpha}\int_{-\infty}^0 (1+y^2)\exp\left(-\frac{y^2}{2\alpha }\right)dy\\ &=\alpha e^{-1/\alpha}(\alpha+1)\sqrt{\frac{\alpha\pi}{2}}. \end{align*}$$


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