A closed form expression for $\int_0^{\infty} (t^2+t^4)^n e^{-t^2-t^4}\,dt$

Question

I was doing some computations for research purposes, which led me to this integral:

$$I(n) = \int_0^{\infty} (t^2+t^4)^n e^{-t^2-t^4}\,dt.$$

This is very suggestively written so as to employ a parametric differentiation technique as so:

$$\left(\frac{\partial^n}{\partial \alpha^n}\right)\int_0^{\infty}e^{-\alpha(t^2+t^4)}\,dt.$$

This integral has a nice, closed form expression:

$$\int_0^{\infty} e^{-\alpha(t^2+t^4)}\,dt = \frac{1}{4} e^{\frac{\alpha}{8}}K_{\frac{1}{4}}\left(\frac{\alpha}{8}\right),$$

where $K_{\nu}$ is the modified Bessel function of the second kind. From here, I would have to employ $n$ differentiations which would be pretty messy to work out due to product rule. $n$ applications of product rule does have a nice combinatorial expression but it is far from explicit. Moreover, Bessel functions can get pretty complicated after differentiating so this seems like a bad approach.

Instead I decided to run some examples on Mathematica and computed the first 22 of these and noticed a very surprising pattern. In what follows $I_{\nu}$ is the modified Bessel function of the first kind.

$$I(0) = \frac{1}{4} e^{\frac{1}{8}}K_{\frac{1}{4}}\left(\frac{1}{8}\right)$$

$$I(1) = \frac{1}{32} e^{\frac{1}{8}}\left(K_{\frac{1}{4}}\left(\frac{1}{8}\right) + K_{\frac{3}{4}}\left(\frac{1}{8}\right)\right)$$

$$I(2) = \frac{3}{128\sqrt{2}} e^{\frac{1}{8}}\pi \left(3 I_{-\frac{1}{4}} \left(\frac{1}{8}\right) + I_{\frac{1}{4}}\left(\frac{1}{8}\right) - I_{\frac{3}{4}}\left(\frac{1}{8}\right) + I_{\frac{5}{4}}\left(\frac{1}{8}\right)\right)$$

$$I(3) = \frac{1}{256\sqrt{2}} e^{\frac{1}{8}}\pi \left(39 I_{-\frac{1}{4}} \left(\frac{1}{8}\right) + 17 I_{\frac{1}{4}}\left(\frac{1}{8}\right) - 14 I_{\frac{3}{4}}\left(\frac{1}{8}\right) + 14 I_{\frac{5}{4}}\left(\frac{1}{8}\right)\right)$$

$$I(4) = \frac{1}{2048\sqrt{2}} e^{\frac{1}{8}} \pi \left(1029 I_{-\frac{1}{4}} \left(\frac{1}{8}\right) + 367 I_{\frac{1}{4}}\left(\frac{1}{8}\right) - 349 I_{\frac{3}{4}}\left(\frac{1}{8}\right) + 349 I_{\frac{5}{4}}\left(\frac{1}{8}\right)\right)$$

$$I(5) = \frac{9}{8192\sqrt{2}} e^{\frac{1}{8}} \pi \left(1953 I_{-\frac{1}{4}} \left(\frac{1}{8}\right) + 619 I_{\frac{1}{4}}\left(\frac{1}{8}\right) - 643 I_{\frac{3}{4}}\left(\frac{1}{8}\right) + 643 I_{\frac{5}{4}}\left(\frac{1}{8}\right)\right)$$

$$I(6) = \frac{1}{16384\sqrt{2}} e^{\frac{1}{8}} \pi \left(185157 I_{-\frac{1}{4}} \left(\frac{1}{8}\right) + 53131 I_{\frac{1}{4}}\left(\frac{1}{8}\right) - 59572 I_{\frac{3}{4}}\left(\frac{1}{8}\right) + 59572 I_{\frac{5}{4}}\left(\frac{1}{8}\right)\right)$$

Repeat ad nauseum. Each of the terms in the denominator seems to be a power of $2$, the third and fourth terms seem to have the same coefficient (modulo a sign) and the signs are $+$, $+$, $-$, $+$. The "nice" output seems to suggest to me that there is a closed-form expression for $I(n)$ in general but I haven't the slightest clue as to how to come up with it. Can anyone shed some light on the matter?

A PDF with more expressions can be found here. (Mathematica output.)

Answer 1

Let us make the change of variables $t=\sinh \frac{x}{4}$. Since $$t^2+t^4=\frac{\sinh^2\frac{x}{2}}{4}=\frac{\cosh x-1}{8}$$ and $dt=\frac14 \cosh\frac{x}{4}dx$, the initial integral can be rewritten as \begin{align} I(n)&=\frac{e^{\frac18}}{4^{n+1}}\int_0^{\infty}\cosh\frac{x}{4}\sinh^{2n}\frac{x}{2}e^{-\frac18\cosh x}dx=\\ &=\frac{e^{\frac18}}{2^{4n+3}}\int_{-\infty}^{\infty}e^{\frac{x}{4}}\left(e^{\frac{x}{2}}-e^{-\frac{x}{2}}\right)^{2n}e^{-\frac18\cosh x}dx=\\ &=\frac{e^{\frac18}}{2^{4n+3}}\sum_{k=0}^{2n} \left(-1\right)^k {2n \choose k}\int_{-\infty}^{\infty}e^{(n-k+\frac{1}{4})x-\frac18\cosh x}dx=\\ &=\frac{e^{\frac18}}{2^{4n+2}}\sum_{k=0}^{2n} \left(-1\right)^k {2n \choose k} K_{n-k+\frac14}\left(\frac18\right). \end{align} At the last step, I use the integral representation $\displaystyle K_{\nu}(r)=\frac12\int_{-\infty}^{\infty}e^{-r\cosh x+\nu x}dx$.

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Added: Using the recurrence relation $K_{\nu+1}(r)-K_{\nu-1}(r)=\frac{2\nu}{r}K_{\nu}(r)$, the last sum (1) can be expressed in terms of rational multiples of only two Macdonald functions, for example, $K_{\frac14}\left(\frac18\right)$ and $K_{\frac54}\left(\frac18\right)$. However it seems unlikely that the corresponding coefficients can be written in a nice way.

Answer 2

I would think that the parametric differentiation technique is probably easier to carry through:

$$\frac{\partial}{\partial r}K_{\nu}(r)= -(1/2)(K_{\nu-1}(r) + K_{\nu+1}(r))......(1)$$

EDIT: From the answer by O.L., we know that:

$$K_{\nu+1}(r)-K_{\nu-1}(r)=\frac{2\nu}{r}K_{\nu}(r)......(2)$$

Thus we can get rid of $K_{\nu-1}(r)$ and obtain: $$\frac{\partial}{\partial r}K_{\nu}(r)= -K_{\nu+1}(r) +(1/2)\frac{2\nu}{r}K_{\nu}(r)......(3)$$

Therefore, as O.L. pointed out, the final result can be expressed as a linear combination of $K_{1/4}(1/8)$ and $K_{5/4}(1/8)$.

Answer 3

Not a full answer, but a partial one:
Notice that the powers of two in the denominator are increasing in periodic repetitions of 3, 2, 1. That is, in this case the exponents on the 2 are 2, 5, 7, 8, 11, 13, 14, 17, 19, 20.... This is sequence A047268 on the OEIS website, defined as the sequence of numbers congruent to {1, 3, 5} mod(6).

I fiddled around with the other coefficients, however nothing useful turned out (no known sequences with the numbers themselves or their differences, products, sums...). If you absolutely need a closed form expression for it, then you could just define three new sequences $S_1$, $S_2$ and $S_3$ as the coefficients of the first three modified Bessel functions of the first kind in your expression. Then the closed form expression would be:

$I(n) = 2^{-\delta_{n + 2}}e^{\frac18}(K_{\frac14}(\frac18)+nK_{\frac34}(\frac18))$ $ n = 0,1$

$I(n) = 2^{-\frac{\delta_{n + 2}}{2}}e^{\frac18}\pi(S_1I_{-\frac{1}{4}}(\frac18) + S_2I_{\frac{1}{4}}(\frac18) - S_3I_{\frac{3}{4}}(\frac18) + S_3I_{\frac{5}{4}}(\frac18))$ $ n \geq 2$.

Where,
$\delta_n$ is sequence A047268,
$S_1, S_2, S_3$ are explicitly given by you. Note that the coefficients are not simplified by being factored out, as is for example in I(2) where 3 is factored out. You need to multiply them back in to get the sequences so there will always be a coefficient of 1 outside the numerator.

Finally, if you really want to get compact, you can write the second expression as such:
$I(n) = 2^{-\frac{\delta_{n + 2}}{2}}e^{\frac18}\pi\sum\limits_{k = 1}^4 S_kI_{\frac{2k - 3}{2}}(\frac18)$.

Where $\delta_n$ and $S_1, S_2, S_3$ are defined as above, and $S_4 = -S_3$ (it appears implicitly in the summation).

Hope this helps!