Integral $\int_0^\infty \ln x\,\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) \frac{x^4+3\alpha x^2- 1}{x^6}dx$

Question

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

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.

I already asked you this question to which you never answered : are you planning to produce a textbook with "nice" integrals ? — Claude Leibovici, May 03 '14 at 06:20
@ClaudeLeibovici I did respond, I apologize you never saw it. No this is just for everybody on math stack. I am integrals, so I contribute integrals for us all. I do not plan on making a textbook with nice integrals, those already exist! — Jeff Faraci, May 03 '14 at 06:21
@nbubis I have many variations of these kinds....Once you can solve one, you try and generalize. I used to love these as a kid, i'm digging them all out again — Jeff Faraci, May 03 '14 at 06:22
@Integrals. Thanks for answering. May I ask you where you find these integrands and results ? — Claude Leibovici, May 03 '14 at 06:23
@ClaudeLeibovici I have books of my notes as a math student, we used to study integration very seriously. I have about 10 variations of this integral in my notes from the 70/80's. Now I am old. I do check the results numerically — Jeff Faraci, May 03 '14 at 06:25
To evaluate the last three integrals you can use the technique in my answer. — Mhenni Benghorbal, May 03 '14 at 06:44
@ClaudeLeibovici if there are infinitely many primes, surely the number 72 is quite small. Thank you for your answer — Jeff Faraci, May 03 '14 at 14:46
@ChenWang Yes sorry typo from the other post , it should be $3\alpha x^2$ — Jeff Faraci, May 05 '14 at 13:55

score 7 · Accepted Answer · answered May 05 '14 at 10:08

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

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

Integral $\int_0^\infty \ln x\,\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) \frac{x^4+3\alpha x^2- 1}{x^6}dx$

1 Answers1