What is $\int_0^{\infty}e^{-x^{1.5}+\theta x}dx=?$

Question

Is there a way to obtain an expression for $$\int_0^{\infty}e^{-x^{1.5}+\theta x}dx=?$$

If $\theta=0$, we know the above is the same as $\frac{\Gamma(2/1.5)}{\Gamma(1/1.5)}$ from a generalized Gamma function. Also, I am interested in $$\int_0^{\infty}xe^{-x^{1.5}+\theta x}dx=?$$ or, $$\int_0^{\infty}\sqrt xe^{-x^{1.5}+\theta x}dx=?$$

It seems the 2nd integral above has to do with the generalized normal density, but coudn't figure out the exact connection. I am particularly interested in the limiting behavior of the above integrals when $\theta$ gets very large?

Any help or intuitions would be very appreciated!

Answer 1

For the first integral, Maple is giving the following result

when θ gets large we obtain

and for very large values of θ is reduced to

Using the approximation

we obtain the asymptotic result

The following picture shows the behavior of the integral respect to θ

Answer 2

I think this integral can be approximated for large $\theta$ using a variant of saddle-point integration. I should preface this by noting that it's been a long time since I've had to do an integral this way, so I'll be (roughly) following the technique in these notes. I welcome any corrections to/complaints about the derivation below.

Change variables such that $$ w = \sqrt{2 \theta}\left(\sqrt{x} - \frac{2\theta}{3}\right). $$ The integral as a function of $\theta$ is then \begin{align} I(\theta) &= \int_0^\infty \exp (-x^{3/2} + \theta x) dx \\ &= \int_{-(2\theta)^{3/2}/3}^\infty \exp \left[ \frac{4 \theta^3}{27} - \frac{w^2}{2} - \frac{w^3}{(2 \theta)^{3/2}}\right] \frac{2}{\sqrt{2\theta}}\left(\frac{w}{\sqrt{2 \theta}} + \frac{2\theta}{3} \right)dw \\ &= \frac{2 \sqrt{2 \theta}}{3} e^{4 \theta^3/27} \int_{-(2\theta)^{3/2}/3}^\infty e^{-w^2/2} \left(1 + \frac{3 w}{(2 \theta)^{3/2}} \right) \left[ 1 - \frac{w^3}{(2 \theta)^{3/2}} + \frac{w^6}{2 (2\theta)^3} + \dots \right] dw \end{align} where in the last term we have expanded $e^{-w^3/(2 \theta)^{3/2}}$ about $w = 0$.

Now, the integral will largely be dominated by the Gaussian factor $e^{-w^2/2}$ over the region of integration. This means** that we can extend the range of integration to the entire real axis, as the relative error introduced this way is small. Thus, the integral can be approximated as $$ I(\theta) \approx \frac{2 \sqrt{2 \theta}}{3} e^{4 \theta^3/27} \int_{-\infty}^\infty e^{-w^2/2} \left[1 + \frac{3w - w^3}{2^{3/2}} \theta^{-3/2} + \frac{w^6/2- 3w^4}{8\theta^3} + \mathcal{O}(\theta^{-9/2}) \right] $$ Since the second term is square brackets is an odd polynomial, it vanishes in the integration. In fact, the $\mathcal{O}(\theta^{-9/2})$ term in the expansion will be an odd polynomial as well; the next non-vanishing integral will be $\mathcal{O}(\theta^{-6})$. We are left with the result that $$ \boxed{I(\theta) \approx \frac{4 \sqrt{\pi \theta} }{3} e^{4 \theta^3/27} \left( 1 - \frac{3}{16}\theta^{-3} + \mathcal{O}(\theta^{-6})\right).} $$

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** NOTE: This is the part I'm least sure about. In the standard sort of saddle-point integration, one has the lower bound of integration of the integral approaching $-\infty$ as the square root of some parameter; here, it's going as the 3/2 power of that parameter. My intuition is that this actually means that the error introduced by extending the integral to the real line is lower than that in the standard case, but there may be a subtlety I'm missing.