Integrate squared gaussians I need to compute the integral
$$
\int_{-\infty}^\infty e^{-a^4 - (1-a-b)^4} da
$$
for which it seems there is no analytic solution. Trying with Mathematica also gives no solution.
Do you have any suggestion, or possible method to solve this? (e.g. smart change of variables, ...)
 A: First let's rewrite the integral a bit. Introducing $1-b=2c$, we have:
\begin{align}
&\int_{-\infty}^\infty e^{-a^4-(1-a-b)^4}\mathrm{d}a=\int_{-\infty}^\infty e^{-a^4-(2c-a)^4}\mathrm{d}a=\int_{-\infty}^\infty e^{-(c+a-c)^4-(c+c-
a)^4}\mathrm{d}a=\\
&=\int_{-\infty}^\infty e^{-(c+x)^4-(c-x)^4}\mathrm{d}x=
=\int_{-\infty}^{\infty}e^{-2(x^4+6c^2x^2+c^4)}\mathrm{d}x=e^{16c^4}\int_{-\infty}^{\infty}e^{-2(x^2+3c^2)^2}\mathrm{d}x
\end{align}
We now want to find the last integral. Wolfram Alpha gives
$$
\int_{-\infty}^\infty e^{-2(x^2+y^2)^2}\mathrm{d}x=\frac{|y|}{\sqrt{2}}e^{-y^4}K_{\frac{1}{4}}(y^4)
$$
Here $K$ is a modified Bessel function of the second kind. Plug in $y^2=3c^2$ and then $2c=1-b$ to get the original integral expressed as a function of $b$:
$$
\int_{-\infty}^\infty e^{-a^4-(1-a-b)^4}\mathrm{d}a=\sqrt{\frac{3}{8}}\,\,|b-1|\,\,e^{\frac{7}{16}(b-1)^4}\,K_{\frac{1}{4}}\!\!\left(\frac{9}{16}(b-1)^4\right)
$$
If $b=1$ the integral is easily expressible in terms of the Gamma function and this result agrees with the one found from the general result by taking the limit $b\rightarrow 1$.
I invite anyone to try to calculate the last integral by hand and post a derivation. I suspect it has something to do with the substitution $x\rightarrow y\sinh t$ but I didn't get anywhere.  
EDIT: I managed to calculate the integral.
\begin{align}
\int_{-\infty}^\infty e^{-2(x^2+y^2)^2}\mathrm{d}x=2\int_0^\infty e^{-8((x/\sqrt{2})^2+y^2/2)^2}\mathrm{d}x=2\sqrt{2}\int_0^\infty e^{-8(x^2+y^2/2)^2}\mathrm{d}x
\end{align}
Now make the substitution $x=|y|\sinh (t/4)$.
\begin{align}
2\sqrt{2}\int_0^\infty e^{-8(x^2+y^2/2)^2}\mathrm{d}x=\frac{2\sqrt{2}}{4}|y|\int_0^\infty e^{-8y^4(\sinh(t/4)^2+1/2)^2}\cosh(t/4)\mathrm{d}t
\end{align}
Using appropriate identities, one can find that $$\left(\sinh^2\left(\frac{t}{4}\right)+\frac{1}{2}\right)^2=\frac{1}{8}(1+\cosh(t))$$
Therefore:
$$
\frac{2\sqrt{2}}{4}|y|\int_0^\infty e^{-8y^4(\sinh(t/4)^2+1/2)^2}\cosh(t/4)\mathrm{d}t=
\frac{|y|}{\sqrt{2}}\int_0^\infty e^{-y^4(\cosh(t)+1)}\cosh(t/4)\mathrm{d}t
$$
One definition of $K$ is: 
$$
K_{\alpha}(x)=\int_0^\infty e^{-x\cosh t}\cosh(\alpha t)\mathrm{d}t
$$
So:
$$
\frac{|y|}{\sqrt{2}}\int_0^\infty e^{-y^4(\cosh(t)+1)}\cosh(t/4)\mathrm{d}t=\frac{|y|}{\sqrt{2}}e^{-y^4}K_{\frac{1}{4}}(y^4)
$$
as wanted.
