Evaluating $\int_{0}^{\infty} \sqrt{x} e^{-\left(4x + \frac{9}{x}\right)} \, dx$

Question

I have attempted solving this integral with no success. I would really appreciate a hint, and not the full solution. If you know of multiple ways, I would also be really interested in those.

$$\int_{0}^{\infty} \sqrt{x} e^{-\left(4x + \frac{9}{x}\right)} \, dx$$

edit:

Here is what I have tried so far. In order to keep it as brief as possible I will only list the general tools I tried using, as a full list of everything I have tried would be rather long.

Using the power series of exp(x), however it results in $$\sum_{n=0}^\infty \frac{(-1)^n}{n!} \int_{0}^\infty \sqrt{x} \left(4x + \frac{3}{x}\right)^n \, dx$$ which is not very useful.

Subsituting $$\sqrt{x} = w$$ $$dw = \frac{dx}{2w}$$ gives $$2e^{12} \int_{0}^\infty w^2 e^{\left(2w + \frac{3}{w}\right)^2} dw$$ which is also not very useful.

Trying to parametrize the integral as $$\psi(\alpha) = \int_{0}^\infty \sqrt{x} e^\left(-4x - \frac{\alpha}{x}\right) \, dx$$ to get $$\frac{d\psi}{d\alpha} = \int_{0}^\infty xe^{-4x - \frac{\alpha}{x}}dx$$ which is also not very useful.

And using the above mentioned together in various ways. I feel like I must be missing something vital here.

Answer 1

After $x=t^2$, the given integral is $$ 2\int_0^\infty t^2\exp(-4t^2-9/t^2)\,dt=2F(2,3), $$ where $F(a,b)=-\dfrac1{2a}\dfrac{\partial}{\partial a}G(a,b)$ and $$ G(a,b)=\int_0^\infty\exp(-a^2 t^2-b^2/t^2)\,dt=\frac{\sqrt\pi}{2a}e^{-2ab} $$ (for $a,b>0$) is known. Thus $F(a,b)=\dfrac{\sqrt\pi}{4a^3}(1+2ab)e^{-2ab}$ and the answer is $\dfrac{13\sqrt\pi}{16e^{12}}$.

Answer 2

Let's use the bessel function to calculate the integral.

$$I = \int_{0}^{\infty} \sqrt{x} e^{-\left(4x + \frac{9}{x}\right)} \, dx$$ we can use the substitution $ x = t^2 $. Then, we have these things.

$$dx = 2t \, dt,$$ and the limits of integration remain the same since as $ x $ goes from $ 0 $ to $ \infty $, $ t $ also goes from $ 0 $ to $ \infty $. Thus, we can rewrite the integral as: $$I = \int_{0}^{\infty} \sqrt{t^2} e^{-\left(4t^2 + \frac{9}{t^2}\right)} \cdot 2t \, dt = 2 \int_{0}^{\infty} t^2 e^{-\left(4t^2 + \frac{9}{t^2}\right)} \, dt.$$ Now, we simplify the exponent: $I = 2 \int_{0}^{\infty} t^2 e^{-4t^2} e^{-\frac{9}{t^2}} \, dt.$ Next, we can use the substitution $ u = t^2 $, which gives $ du = 2t \, dt $ or $ dt = \frac{du}{2\sqrt{u}} $. The limits remain the same. $I = 2 \int_{0}^{\infty} u e^{-4u} e^{-\frac{9}{u}} \cdot \frac{du}{2\sqrt{u}} = \int_{0}^{\infty} u^{1/2} e^{-4u - \frac{9}{u}} \, du.$ Now, we can recognize that this integral can be evaluated using the formula for the integral of this form. $ \int_{0}^{\infty} x^{\nu - 1} e^{-\beta x - \frac{\gamma}{x}} \, dx = 2 \left( \frac{\gamma}{\beta} \right)^{\nu/2} K_{\nu}(2\sqrt{\beta \gamma}), $ where $ K_{\nu} $ is the modified Bessel function of the second kind. In our case, we have: $ \nu = \frac{3}{2} $ $ \beta = 4 $ $ \gamma = 9 $

Thus, we can compute like under. $ I = 2 \left( \frac{9}{4} \right)^{3/4} K_{\frac{3}{2}}(6). $ The modified Bessel function $ K_{\frac{3}{2}}(x) $ can be computed as this. $ K_{\frac{3}{2}}(x) = \sqrt{\frac{\pi}{2x}} e^{-x} \left( 1 + \frac{1}{x} \right). $ Substituting $ x = 6 $. $ K_{\frac{3}{2}}(6) = \sqrt{\frac{\pi}{12}} e^{-6} \left( 1 + \frac{1}{6} \right) = \sqrt{\frac{\pi}{12}} e^{-6} \cdot \frac{7}{6}. $ Now substituting back into our expression for $ I $: $I = 2 \left( \frac{9}{4} \right)^{3/4} \cdot \sqrt{\frac{\pi}{12}} e^{-6} \cdot \frac{7}{6}. $Calculating $ \left( \frac{9}{4} \right)^{3/4} = \frac{27}{8}$, we have: $I = \frac{27}{8} \cdot \sqrt{\frac{\pi}{12}} \cdot e^{-6} \cdot \frac{7}{6}.$ Simplifying further: $I = \frac{27 \cdot 7}{48} \cdot \sqrt{\frac{\pi}{12}} \cdot e^{-6} = \frac{189}{48} \cdot \sqrt{\frac{\pi}{12}} \cdot e^{-6}.$ Finally, simplifying $ \sqrt{\frac{\pi}{12}} = \frac{\sqrt{\pi}}{2\sqrt{3}} $.

$I = \frac{189 \sqrt{\pi}}{96 \sqrt{3}} e^{-6}.$ Thus, the final result is $ I = \frac{13 \sqrt{\pi}}{16} e^{-12}. $

Answer 3

Using $\sqrt x\mapsto x $ yields $$ \int_0^{\infty} \sqrt{x} e^{-\left(4 x+\frac{9}{x}\right)} d x = 2 \int_0^{\infty} x^2 e^{-\left(4 x^2+\frac{9}{x^2}\right)} d x=I(9) $$ where $$ I(a)=2 \int_0^{\infty} x^2 e^{-\left(4 x^2+\frac{a}{x^2}\right)} d x $$ Differentiating $I(a)$ w.r.t. $a$ brings us $$ \begin{aligned} I^{\prime}(a) & =-2 \int_0^{\infty} e^{-\left(4 x^2+\frac{a}{x^2}\right)} d x \\ & =-\frac{\sqrt{\pi}}{2} e^{-4 \sqrt{a}} \end{aligned} $$ where the last answer using the result in the post.

Integrating back from $a=0$ to $9$ gives $$ I(9)-I(0)=-\frac{\sqrt{\pi}}{2} \int_0^9 e^{-4 \sqrt{a}} d a \stackrel{IBP}{=} \frac{\sqrt{\pi}}{16}\left(\frac{13}{e^{12}}-1\right) $$ Via integration by parts, we have $$ \begin{aligned} I(0) & =2 \int_0^{\infty} x^2 e^{-4 x^2} d x \\ & =-\frac{1}{4} \int_0^{\infty} x d\left(e^{-4 x^2}\right) \\ & =\frac{1}{4} \int_0^{\infty} e^{-4 x^2} d x \\ & =\frac{\sqrt{\pi}}{16} \end{aligned} $$ Hence we can conclude that $$ \boxed{\int_0^{\infty} \sqrt{x} e^{-\left(4 x+\frac{9}{x}\right)} d x =\frac{13 \sqrt{\pi}}{16 e^{12}}} $$

Answer 4

Alternative approach, using linear combination instead of ODE. First, note $$ a^2x^2+\frac{b^2}{x^2} = ab\left(\frac{ax^2}{b}+\frac{b}{ax^2}\right) = ab\left(\sqrt{\frac{a}{b}}x-\sqrt{\frac{b}{a}}\frac1{x}\right)^2+2ab $$ we change the original integral into canonical form of Gaussian integral, let $x\mapsto x^2$ and $x\mapsto\sqrt{\frac3{2}}x$, which gives $$ \begin{aligned} I &=\int_{0}^{\infty}\sqrt{x}\exp\left\{-\left(4x+\frac9{x}\right)\right\} \,\mathrm{d}x = 2\int_{0}^{\infty}x^2\exp\left\{-\left(4x^2+\frac9{x^2}\right)\right\} \,\mathrm{d}x \\ &=2\int_{0}^{\infty}x^2\exp\left\{-6\left(\frac{2x^2}{3}+\frac3{2x^2}\right)\right\} \,\mathrm{d}x =\frac3{e^{12}}\sqrt{\frac3{2}}\int_{0}^{\infty}x^2\exp\left\{-6\left(x-\frac1{x}\right)^2\right\} \,\mathrm{d}x \end{aligned} $$ note if we let $x\mapsto1/x$, every canonical Gaussian integral has its reverse equivalent, like this one $$ I = \frac3{e^{12}}\sqrt{\frac3{2}}\int_{0}^{\infty}x^2\exp\left\{-6\left(x-\frac1{x}\right)^2\right\} \,\mathrm{d}x = \frac3{e^{12}}\sqrt{\frac3{2}}\int_{0}^{\infty}\frac1{x^4}\exp\left\{-6\left(x-\frac1{x}\right)^2\right\} \,\mathrm{d}x $$ introduce such a technique, denote $$ \begin{aligned} J &= \frac3{e^{12}}\sqrt{\frac3{2}}\int_{0}^{\infty}\exp\left\{-6\left(x-\frac1{x}\right)^2\right\} \,\mathrm{d}x = \frac3{e^{12}}\sqrt{\frac3{2}}\int_{0}^{\infty}\frac1{x^2}\exp\left\{-6\left(x-\frac1{x}\right)^2\right\} \,\mathrm{d}x \\ &=\frac3{2e^{12}}\sqrt{\frac3{2}}\int_{0}^{\infty}\left(x+\frac1{x^2}\right)\exp\left\{-6\left(x-\frac1{x}\right)^2\right\} \,\mathrm{d}x \\ &=\frac3{2e^{12}}\sqrt{\frac3{2}}\int_{0}^{\infty}\exp\left\{-6\left(x-\frac1{x}\right)^2\right\} \,\mathrm{d}\left(x-\frac1{x}\right) \\ &=\frac3{2e^{12}}\sqrt{\frac3{2}}\int_{-\infty}^{\infty}e^{-6x^2} \,\mathrm{d}x = \frac{3\sqrt{\pi}}{4e^{12}} \end{aligned} $$ where recall the basic Gaussian integral $$ \int_{-\infty}^{\infty}e^{-ax^2} \,\mathrm{d}x = \sqrt{\frac{\pi}{a}} $$ for higher order integral, note that $$ \left(1+\frac1{x^2}\right)\left(1-\frac1{x}\right)^2 = x^2+\frac1{x^4} -\left(1+\frac1{x^2}\right) $$ we have linear combination $$ \begin{aligned} 2I-2J &= \frac3{e^{12}}\sqrt{\frac3{2}}\int_{0}^{\infty}\left(1+\frac1{x^2}\right)\left(1-\frac1{x}\right)^2\exp\left\{-6\left(x-\frac1{x}\right)^2\right\} \,\mathrm{d}x \\ &= \frac3{e^{12}}\sqrt{\frac3{2}}\int_{-\infty}^{\infty}x^2e^{-6x^2} \,\mathrm{d}x = \frac{\sqrt{\pi}}{8e^{12}} \end{aligned} $$ hence $$ I=\int_{0}^{\infty}\sqrt{x}\exp\left\{-\left(4x+\frac9{x}\right)\right\} \,\mathrm{d}x=\frac{\sqrt{\pi}}{16e^{12}} + J = \frac{13\sqrt{\pi}}{16e^{12}} $$