How do I evaluate $$\displaystyle\int^{\infty}_0 \exp\left[-\left(4x+\dfrac{9}{x}\right)\right] \sqrt{x}\;dx?$$

To my knowledge the following integral should be related to the Gamma function.

I have tried using the substitution $t^2 = x$, and I got $$ 2e^{12}\displaystyle \int^{\infty}_0 \exp\left[-\left(2t + \dfrac{3}{t}\right)^2\right] t^2 \; dt $$ after substitution. But it seems like I can do nothing about this integral anymore. Can anyone kindly give me a hint, or guide me to the answer?

  • $\begingroup$ Integration by parts? $\endgroup$ – Toby Mak Aug 20 '19 at 11:40
  • $\begingroup$ The result should be $$\frac{13 \sqrt{\pi }}{16 e^{12}}$$ $\endgroup$ – Dr. Sonnhard Graubner Aug 20 '19 at 12:18
  • 3
    $\begingroup$ See this answer: math.stackexchange.com/a/1820055/269624, it may help $\endgroup$ – Yuriy S Aug 20 '19 at 12:37
  • $\begingroup$ @Zacky Oh sorry, I won't do it again $\endgroup$ – Max Wong Aug 23 '19 at 15:51

(I started an answer involving only "plain computations", but was not quick enough, maybe it is time now to complete and submit, rather then removed the typed formulas and quit the post. I am posting an alternative solution in the hope it looks simpler from some point of view, although there is a lot to be typed.)

We have to compute the integral: $$ \begin{aligned} J&= \int_0^{\infty} \exp\left(-\left(4x+\frac{9}{x}\right)\right) \; \sqrt{x}\;dx \\ &\qquad\text{Substitution, so formally: $t=2\sqrt x$, $t^2=4x$, $x=t^2/4$, $dx=\frac 12t\; dt$} \\ &= \int_0^{\infty} \exp\left(-\left(t^2+\frac{36}{t^2}\right)\right) \; \frac 12 t\cdot \frac 12t\; dt \\ &= \frac 14 e^{-12} \underbrace{ \int_0^{\infty} \exp\left(-\left(t-\frac{6}t\right)^2\right) t^2\; dt}_{\text{Notation: }K} \\[3mm] &\qquad\text{ and we want to show the above is equal to...} \\ &\overset{(?)}= \frac 14 e^{-12}\cdot\frac14\cdot 13\sqrt \pi\ . \\[3mm] &\qquad\text{ So we consider the integral...} \\ K&= \int_0^{\infty} \exp\left(-\left(t-\frac{6}t\right)^2\right) t^2\; dt \\ &\qquad\text{ Substitution $\displaystyle s =t-\frac 6t $, so formally $t^2-st-6=0$,} \\ &\qquad\text{ we use $t=\frac 12(s+\sqrt{s^2+24})$, formally $\displaystyle dt=\frac12 \left(1+\frac s{\sqrt{s^2+24}}\right)\; ds $...} \\ &= \int_{-\infty}^{\infty} e^{-s^2}\cdot \frac 14 (s^2+\color{blue}{2s}\sqrt{s^2+24}+(s^2+24)) \; \frac 12 \left(1+\frac {\color{red}{s}}{\sqrt{s^2+24}}\right)\; ds \\ &\qquad\text{ now expand the parentheses, and ignore the odd part...} \\ &= \frac14\cdot\frac 12 \int_{\Bbb R} e^{-s^2}\;\Big(\ s^2\ +\ (s^2+24)\ + \ \color{blue}{2s}\cdot\color{red}{s} \ \Big) \;ds \\ &= \frac 14\cdot\frac 12 \cdot26\sqrt\pi\ . \\[3mm] &\qquad\text{ Putting all together:} \\ J&= \frac 14 e^{-12} \cdot K \\ &= \frac 14 e^{-12} \cdot \frac 14\cdot\frac 12 \cdot26\sqrt\pi \\ &= \color{magenta}{ \frac {13}{16}\cdot e^{-12} \cdot\sqrt\pi}\ . \end{aligned} $$

Numerical validation, sage code:

sage: J = integral( exp(-4*x-9/x) * sqrt(x), x, 0, oo )
sage: J.n()
sage: ( 13. / 16. * exp(-12) * sqrt(pi) ).n()

sage: var('s');
sage: integral( exp(-s^2) * (s^2+ (s^2+24) + 2*s*s), s, -oo, +oo )

It looks like a tricky integral, however Feynman's trick deals with it nicely. $$I=\int^{\infty}_0 \exp\left(-\left(4x+\dfrac{9}{x}\right)\right) \sqrt{x}dx\overset{\sqrt x\to x}=2\int_0^\infty \exp\left(-\left(4x^2+\frac{9}{x^2}\right)\right)x^2 dx$$ Now consider the following integral: $$I(t)=2\int_0^\infty \exp\left(-\left(4x^2+\frac{t}{x^2}\right)\right)x^2 dx$$ The reason why I'm putting the parameter in that place is because if $x^2$ is simplified then the integral becomes much easier. So let's take a derivative with respect to $t$ in order to get: $$ I'(t)=-2\int_0^\infty \exp\left(-\left(4x^2+\frac{t}{x^2}\right)\right) dx=-\frac{\sqrt \pi}{2}e^{-4\sqrt t}$$ The above result follows using the Cauchy-Schlomilch transformation (see $3.3$).

I think that you are on the right track now and basically the future the steps would be to see that: $$I(0)=\frac{\sqrt \pi}{16}\Rightarrow I=I(9)-I(0)+\frac{\sqrt\pi}{16}=-\frac{\sqrt \pi}2 \int_0^9e^{-4 \sqrt t}dt+\frac{\sqrt{\pi}}{16}=\boxed{\frac{13\sqrt \pi}{16e^2}}$$

  • $\begingroup$ Can you explain the Cauchy-Schlomilch transformation that has been use in your derivation. I do not understand it although I have read the paper x.x $\endgroup$ – Tuong Nguyen Minh Jan 8 at 13:26

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