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I'm new to this forum, and this may be well above my level of understanding, with only high school level of calculus. But I'm very interested in the asymptotic behavior of special functions.

I'm trying to wrap my head around the asymptotic expansion of $\int_{0}^{x}t^{\alpha t}dt$ from this article "https://www.scribd.com/document/34977341/Sophomore-s-Dream-Function" page 6-7. I've tried to do the integration by parts in all ways I can imagine. The most simple of which I get the following:

$\int e^{\alpha x ln(x)} dx= xe^{\alpha x ln (x)} - \int xe^{\alpha x ln(x)}\alpha (ln(x)+1) dx$

so I get how the

$\alpha (ln(x)+1)$

Comes to be, but how it moves to the denominator and how the rest of the expansion comes around I can't figure out. No matter what I do, I end up with something nasty.

Is anyone up to the task, of explaining this as simple and thorough as possible, so even a person with high school calculus could understand?

I see the author of the paper @JJacquelin is active in this forum, It would be delightful if the author themself could answer this question.

Thanks

Edit: The article says the asymptotic expansion is: $\int e^{\alpha x ln(x)} dx \approx \frac{e^{\alpha x ln(x)}}{\alpha (ln(x)+1)} (1+\sum_{n=1}^{n<<\alpha}\frac{1}{\alpha^nx^n}\sum_{j=1}^{n}\frac{A(n,j))}{(1+ln(x))^{n+j}})$

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I'm not the author, but hopefully this will tide you over. (I can't access scribd, but I did find this link.) What you should do is integrate by parts as follows: recall that $(e^{\alpha x \ln x})' = \alpha (\ln x + 1) e^{\alpha x \ln x}. $ Dividing both sides by $\alpha(\ln x+1)$, $e^{\alpha x \ln x} =(e^{\alpha x \ln x})' \times \frac1{\alpha(\ln x+1)}. $ Then

$$ \int e^{\alpha x \ln x}dx = \int (e^{\alpha x \ln x})' \times \frac1{\alpha(\ln x+1)} dx = \frac{e^{\alpha x \ln x}}{\alpha(\ln x + 1)} - \int e^{\alpha x \ln x} \left(\frac1{\alpha(\ln x+1)} \right)' dx. $$

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  • $\begingroup$ This helped me enough, so I was now able to figure out the whole asymptotic expansion, thanks! $\endgroup$ – Patrick Christensen Nov 11 '19 at 19:18
  • $\begingroup$ @PatrickChristensen glad to have helped! :) Generally if you've gotten the help you need, you click the green tick next to the most helpful answer. You can wait to see if someone else wants to add an answer, or if you want to help other people, you can write out the whole computation as an answer to your own question, and accept that :) $\endgroup$ – Calvin Khor Nov 12 '19 at 5:15

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