How can I evaluate $\int_0^\infty\frac{\sin(x^4)}{x^a(e^x-1)^2}\,\mathrm{d}x$

Question

I can't evaluate this kind of integral.

$$\int_{0}^{\infty}\frac{\sin(x^4)}{x^a(e^x-1)^2}\,\mathrm{d}x$$

whereas: $a\in\mathbb{R}$

Answer 1

$$\int_0^{+\infty} \frac{\sin(x^4)}{x^a (e^x - 1)^2}\ dx = \int_0^{+\infty} \frac{\sin(x^4)}{x^a e^{2x}} \frac{1}{(1 - e^{-x})^2}\ dx = \sum_{k = 0}^{+\infty}(k+1) \int_0^{+\infty} e^{-kx} \frac{\sin(x^4)}{x^a e^{2x}}\ dx$$

By writing

$$\sin(x^4) = \frac{1}{2i}\left(e^{ix^4} - e^{-ix^4}\right)$$

we can arrange the integral to

$$\sum_{k = 0}^{+\infty}(k+1)\frac{1}{2i} \int_0^{+\infty} e^{-x(k+2)}\left(\frac{e^{ix^4} - e^{-ix^4}}{x^a}\right)\ dx $$

Defining for simplicity $s = k+2$:

$$\sum_{s = 2}^{+\infty}(s-1)\frac{1}{2i} \int_0^{+\infty} e^{-xs}\left(\frac{e^{ix^4} - e^{-ix^4}}{x^a}\right)\ dx $$

At this point, methods may differ according to ideas.

• Use the stationary phase method

• Use other numerical methods

• See the round bracket as a $F(x)$ function, which'd turn the integral into a (HARD) Laplace Transform

• Use Taylor Series for the second Exponential term, the negative one

• Some counter integration (?)

• $\ldots$ Use your fantasy, but stay on $\mathbb{R}$

And in any case: beware of $a$, it may make things to diverge...