How one can evaluate the following improper integral \begin{equation} \int_{0}^{\infty} \!\frac{1-e^{-2x^2}}{xe^{x^4}}\mathrm{d}x \end{equation} It seems that one can consider the function \begin{equation} I(\alpha)=\int_{0}^{\infty} \!\frac{1-e^{-\alpha x^2}}{xe^{x^4}}\mathrm{d}x \end{equation} and try to evaluate $I'(\alpha)=\int_{0}^{\infty} xe^{-\alpha x^2-x^4}\mathrm{d}x$. But this approach did not work for me.

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    $\begingroup$ Complete the square in the exponent to get $-\alpha x^2-x^4=\frac{\alpha^2}4-(x^2+\alpha)^2$, and now you can use a substitution to compute $I'(\alpha)$. $\endgroup$ – user170231 Feb 26 at 22:24
  • $\begingroup$ If you want to evaluate $\int_0^{\infty} xe^{-\alpha x^2 - x^4}\, dx$, you can try a substitution $u = x^2$, so the integral becomes $\frac{1}{2}\int_0^{\infty} e^{-\alpha u - u^2}\, du$. Then you can try completing the square in the exponent with the aim of reducing this to the Gaussian integral. $\endgroup$ – Minus One-Twelfth Feb 26 at 22:24
  • $\begingroup$ @MinusOne-Twelfth There will be the Error function, and then I will need to integrate it. $\endgroup$ – Tzara_T'hong Feb 26 at 22:35

$$ \int_0^\infty \frac{1-e^{-2x^2}}{xe^{x^4}}dx=\int_0^\infty \frac{1-e^{-2u}}{2ue^{u^2}}du =\int_0^\infty e^{-u^2} du \sum_{k=0}^\infty(-1)^k \frac{(2u)^k}{(k+1)!}\\ =\sum_{k=0}^\infty(-1)^k\frac{2^{k-1}\Gamma\left(\frac{k+1}{2}\right)}{(k+1)!} =\frac{\sqrt\pi}2\sum_{n=0}^\infty\frac1{(2n+1)n!}-\sum_{n=0}^\infty\frac{2^n}{(n+1)(2n+1)!!}\\ =\frac{\sqrt\pi}2 {}_1F_1\left(\small\frac12;\small\frac32;1\right)-\frac12 {}_2F_2\left(1,1;\small\frac32,2;1\right). $$

According to WA, the first summand can be represented as: $$ \frac\pi4 \text{Erfi}(1). $$

Further simplification seems to be not possible.


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