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I was told by this OP, $$\int_{0}^{\infty} e^{\large-x^n} \,dx =\Gamma \left(\frac{n+1}{n}\right), \qquad\text{ for $n>1$}.$$ This is via the variable change $t=x^n$:

$$\int_{0}^{\infty} e^{\large-x^n}\,dx=\frac{1}{n}\int_0^{\infty} t^{\large\frac{1}{n}-1}\ e^{-t}\ dt=\frac{1}{n}\Gamma \left(\frac{1}{n}\right)=\Gamma \left(\frac{n+1}{n}\right)$$

However, is it possible to do integral $$I_1=\int_0^{+\infty} e^{\large-x^2-ax^4}\ dx, \qquad\text{ for $a>0$}$$

Or in general, is it possible to do integral $$I_2=\int_0^{+\infty} e^{\large-a_1x-a_2x^2-\cdots-a_nx^n} dx, \qquad\text{ for $a_i>0$}$$


I tried to follow hints from MathFacts

$$4I_1^2=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-x^2-ax^4-y^2-ay^4}dxdy = =\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-(x^2+y^2)-a(x^2+y^2)^2+2ax^2y^2}dxdy$$

change to $(r,\theta)$:

$$4I_1^2 = \int_0^\infty df \, e^{-r^2-ar^4} \int_0^{2\pi} e^{2ar^4\cos^2\theta \sin^2\theta} d\theta$$

Then I'm lost on how to proceed

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  • $\begingroup$ Do $I = \int_{0}^{\infty}e^{-x^2 - ax^4}dx = \int_{0}^{\infty}e^{-y^2 - ay^4}dy$. $\endgroup$ – Mathsource Jul 29 '14 at 6:43
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    $\begingroup$ @MathFacts pls see the question updated. did i get your hint? $\endgroup$ – athos Jul 29 '14 at 7:00
  • $\begingroup$ According to Wolfram Alpha $$ \int_0^{\infty} e^{\large-x^2-ax^4}\ dx=\frac{\sqrt[\large8a]{e}}{4\sqrt{a}}K_{\frac14}\left(\frac{1}{8a}\right), \quad\text{for}\ a\in\mathbb{Z}_+, $$ where $K_n$ is Bessel functions of the second kind. $\endgroup$ – Tunk-Fey Jul 29 '14 at 7:15
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The tables of Laplace transforms are of invaluable help :

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For $\int_0^\infty e^{-x^2-ax^4}~dx$ ,

$\int_0^\infty e^{-x^2-ax^4}~dx$

$=\int_0^\infty e^{-x^2(1+ax^2)}~dx$

$=\int_0^\infty e^{-\left(\frac{\sinh x}{\sqrt a}\right)^2\left(1+a\left(\frac{\sinh x}{\sqrt a}\right)^2\right)}~d\left(\dfrac{\sinh x}{\sqrt a}\right)$

$=\dfrac{1}{\sqrt a}\int_0^\infty e^{-\frac{(\sinh^2x)(1+\sinh^2x)}{a}}\cosh x~dx$

$=\dfrac{1}{\sqrt a}\int_0^\infty e^{-\frac{\sinh^2x\cosh^2x}{a}}\cosh x~dx$

$=\dfrac{1}{\sqrt a}\int_0^\infty e^{-\frac{\sinh^22x}{4a}}\cosh x~dx$

$=\dfrac{1}{\sqrt a}\int_0^\infty e^{-\frac{\cosh4x-1}{8a}}\cosh x~dx$

$=\dfrac{e^\frac{1}{8a}}{\sqrt a}\int_0^\infty e^{-\frac{\cosh4x}{8a}}\cosh x~dx$

$=\dfrac{e^\frac{1}{8a}}{\sqrt a}\int_0^\infty e^{-\frac{\cosh x}{8a}}\cosh\dfrac{x}{4}d\left(\dfrac{x}{4}\right)$

$=\dfrac{e^\frac{1}{8a}}{4\sqrt a}\int_0^\infty e^{-\frac{\cosh x}{8a}}\cosh\dfrac{x}{4}dx$

$=\dfrac{e^\frac{1}{8a}}{4\sqrt a}K_\frac{1}{4}\left(\dfrac{1}{8a}\right)$

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