How to evaluate $\int^{\infty}_{-\infty}e^{-x^4-2x^2-1}dx$? I wanted to compute $\int^{\infty}_{-\infty}e^{-x^4-2x^2-1}dx$
Wolfram Alpha gave me :
$$\int^{\infty}_{-\infty}e^{-x^4-2x^2-1}dx=\frac{1}{\sqrt{2e}}K_{\frac{1}{4}}\left(\frac{1}{2}\right)$$
Where $K_{\frac{1}{4}}(z)$ is the modified Bessel function of the second kind satisfying the differential equation
$$z^2y''+zy'-\left(z^2+\frac{1}{16}\right)y=0$$
and some initial conditions.
My question is : how do you relate the first integral and the Bessel function ? Suppose I had no WolframAlpha, how do I know I have to use Bessel functions ?
 A: As mentioned in a comment, one can start from one of the various integral forms of the modified Bessel functions of the second kind. In the present case, for every $x\gt0$, $$K_n(x)=\int_0^\infty\cosh(nt)\mathrm e^{-x\cosh t}\mathrm dt,$$ in particular, $$K_{1/4}(1/2)=\frac12\int_{-\infty}^\infty\cosh(t/4)\mathrm e^{-\cosh(t)/2}\mathrm dt.$$ The change of variable $$x=\sqrt2\sinh(t/4)$$ yields $\mathrm dx=\sqrt2\cosh(t/4)\mathrm dt/4$ and $x^2+1=2\sinh^2(t/4)+1=\cosh(t/2)$, hence $$\cosh(t)=2\cosh^2(t/2)-1=2(x^2+1)^2-1,$$ which shows that $$K_{1/4}(1/2)=\sqrt2\int_{-\infty}^\infty\mathrm e^{-(x^2+1)^2+1/2}\mathrm dx=\sqrt{2\mathrm e}\int_{-\infty}^\infty\mathrm e^{-(x^2+1)^2}\mathrm dx.$$ Finally, $$\int_{-\infty}^\infty\mathrm e^{-x^4-2x^2-1}\mathrm dx=\int_{-\infty}^\infty\mathrm e^{-(x^2+1)^2}\mathrm dx=\frac{K_{1/4}(1/2)}{\sqrt{2\mathrm e}}.$$ More generally, for every positive $(a,b)$, $$\int_{-\infty}^\infty\mathrm e^{-ax^4-2bx^2}\mathrm dx=\sqrt{\frac{b}{2a}}\mathrm e^{b^2/a}\mathrm e^{-a/2}K_{1/4}(a/2).$$
