Integrate: $\int\frac{1}{\sqrt{x^3}} \cdot e^{-\frac{1}{2} \cdot \frac{(a-bx)^2}{x}} dx$

Asked Feb 28 '20 at 17:43

Active Mar 17 '20 at 22:07

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I've been trying to compute the following integral, but I havent been able to. I suspect that a change of variables would be useful, but none have worked thus far. These include $y = 1/\sqrt{t}$ and other variations but none have worked. Mathematica gives me an answer, but I would like to know how to get to that answer.

$$\int\frac{1}{\sqrt{x^3}} \cdot e^{-\frac{1}{2} \cdot \frac{(a-bx)^2}{x}} dx$$

The answer according to Mathematica is:

$$ \frac{1}{a} \sqrt{\frac{\pi}{2}} e^{-2ab}(-2 + erfc(\frac{a - bt}{\sqrt{2t}}) + e^{2ab}erfc(\frac{a + bt}{\sqrt{2t}})) $$

where $erfc$ is the complementary error function.

edited Feb 28 '20 at 18:24

asked Feb 28 '20 at 17:43

cotton

maybe let U = x^3/2, and try a clever substitution? – Mohammad Athar Feb 28 '20 at 17:46
What is the answer given by Mathematica? – Vincent Feb 28 '20 at 18:03
i updated my question to reflect Mathematicas output – cotton Feb 28 '20 at 18:25
Why not try $y=\frac{a-bx}{\sqrt{2x}}$? – mjw Feb 28 '20 at 18:28
Does this answer your question? Integral of the form $\int_0^\infty \exp{\left(-\frac{(a+bx)^2}{x}\right)}\frac{dx}{\sqrt{x^3}}$ You will need to work with $2x=t$ and $b=-b'$ though. – Zacky Feb 28 '20 at 18:35
@Zacky thank you, that does help. Its difficult to search for integral solutions on stack exchange because im not sure how to refer to them – cotton Feb 28 '20 at 18:39
In this case I searched directly through my answers, but usually approach0 is helpful (just put the integral there). However I might have made a mistake as you're looking to find the indefinite integral not the definite one and it should be marked as a duplicate. Let me know if you need further help. – Zacky Feb 28 '20 at 18:48
@Zacky thank you for your help. I'm a bit unclear about the second substitution in blue in the answer you link. I'm substituting $y = 1/x^2$ and then $dy = -2/x^3 dx$ as you suggest before the blue, but i'm not getting the same intermediate result. Sorry I'm contact you on here. I don't have enough rep on Mathematics to comment on your reply from that post – cotton Feb 28 '20 at 23:15
@Zacky i think i got it now, there is one extra change of variables that is needed to get to blue, i believe – cotton Feb 28 '20 at 23:27
@cotton that was a typo, thanks for catching that. It should have been just $y=1/x\Rightarrow dy=-\frac{1}{x^2}dx$ before the blue integral arrived. – Zacky Feb 28 '20 at 23:40
Perhaps this post https://math.stackexchange.com/q/3583334/515527 is more useful. – Zacky Mar 17 '20 at 22:05

Integrate: $\int\frac{1}{\sqrt{x^3}} \cdot e^{-\frac{1}{2} \cdot \frac{(a-bx)^2}{x}} dx$

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