# Evaluating $\int_{0}^{\infty} e^{-(x^{n}+x)}\, \mathrm{d}x,\quad n>0$

I know from Discussing the Integral of $\exp(-x^n)$ that $$\int_{0}^{\infty} e^{-x^{n}}\mathrm{d}x=\Gamma(1+1/n),\quad n>0.$$

But how to evaluate $$\int_{0}^{\infty}e^{-(x^{n}-x)}\,\mathrm{d}x,\quad n>0?$$

The only substitution i found is $$\text{Let}\quad x=\ln u, \quad \text{then}\quad e^{x}=u, \quad \text{and} \quad \mathrm{d}x=\frac{1}{u}\mathrm{d}u.$$ Then $$\int_{0}^{\infty}e^{-(x^{n}-x)}\,\mathrm{d}x=\int_{1}^{\infty}e^{-(\ln u)^{n}}\,\mathrm{d}u$$ But after this, I am stuck.

Thank you!

• Your title is probably wrong. You seem to mean $\exp(-x^n + x)$, not $\exp(x^n+x)$. – Cameron Williams Jul 7 '18 at 17:56
• Which is the integral? The one in the title which surely diverges if $n>1$ or the one in the question? – Bernard Massé Jul 7 '18 at 17:57
• This integral is going to be a special function which means that there's no really nice way to represent it, and depending on your outlook, the special function representation might be just a different way to write the integral (i.e. it's different notation for the same thing). – Cameron Williams Jul 7 '18 at 17:59
• Sorry, @Cameron Williams. It was a typo. I have corrected it now, – Sofia Fredriksson Jul 7 '18 at 18:01
• The title and the body still don't match though. – Maxim Jul 26 '18 at 13:26

## 2 Answers

Just to add one more nightmare to marty cohen's list.

For $n=6$ $$\frac{\sqrt{\pi }}{12} \, _0F_4\left(;\frac{2}{3},\frac{5}{6},\frac{7}{6},\frac{4}{3};\frac{1}{46656}\right)+\Gamma \left(\frac{7}{6}\right) \, _0F_4\left(;\frac{1}{3},\frac{1}{2},\frac{2}{3},\frac{5}{6};\frac{1}{46656}\right)-$$ $$\frac{1}{9} \Gamma \left(-\frac{2}{3}\right) \, _0F_4\left(;\frac{1}{2},\frac{2}{3},\frac{5}{6},\frac{7}{6};\frac{1}{46656}\right)-\frac{1}{108} \Gamma \left(-\frac{1}{3}\right) \, _0F_4\left(;\frac{5}{6},\frac{7}{6},\frac{4}{3},\frac{3}{2};\frac{1}{46656}\right)-$$ $$\frac{1}{864} \Gamma \left(-\frac{1}{6}\right) \, _0F_4\left(;\frac{7}{6},\frac{4}{3},\frac{3}{2},\frac{5}{3};\frac{1}{46656}\right)+\frac{1}{720} \, _1F_5\left(1;\frac{7}{6},\frac{4}{3},\frac{3}{2},\frac{5}{3},\frac{11}{6};\frac{ 1}{46656}\right)$$ which is $\approx 1.56900$

I may have a mistake somewhere since the values I obtained are $$\left( \begin{array}{cc} n & \int_{0}^{\infty}e^{-(x^{n}-x)}\,dx \\ 2 & 1.73023 \\ 3 & 1.57661 \\ 4 & 1.55602 \\ 5 & 1.55968 \\ 6 & 1.56900 \\ 7 & 1.57924 \\ 8 & 1.58899 \\ 9 & 1.59786 \\ 10 & 1.60582 \\ 11 & 1.61291 \\ 12 & 1.61924 \end{array} \right)$$ showing a minimum

• For sure, this is given by a CAS ! – Claude Leibovici Jul 8 '18 at 7:22
• When $n\to \infty$ define integral is : $e-1$ – Mariusz Iwaniuk Jul 8 '18 at 7:49
• @MariuszIwaniuk. This is fine ! SInce we know the value for $n=3$, there is a minimum somewhere ! I just feel stupid (once more). Cheers. – Claude Leibovici Jul 8 '18 at 8:15

Wolfy gives increasingly complicated expressions for increasing values of $n$. The complexity is the same whether the exponent has $x$ or $-x$.

For $n=2$:

$\int_0^∞ e^{-(x^2 - x)} dx = (1/2) e^{1/4} \sqrt{π} (erf(1/2) + 1)≈1.73023$

For $n=3$:

$\int_0^∞ e^{-(x^3 - x)} dx = (1/18) (3 _1 F_2(1;4/3, 5/3;1/27) + 4 3^{2/3} π Bi(1/3^{1/3}))≈1.57661$

($Bi$ is the Airy $Bi$ function)

For $n=5$:

$\int_0^∞ e^{-(x^5 - x)} dx = (1/120) (_1 F_4(1;6/5, 7/5, 8/5, 9/5;1/3125) + 4 (30 Γ(6/5) _0 F_3(;2/5, 3/5, 4/5;1/3125) + 6 Γ(2/5) _0 F_3(;3/5, 4/5, 6/5;1/3125) + 5 Γ(8/5) _0 F_3(;4/5, 6/5, 7/5;1/3125) + Γ(4/5) _0 F_3(;6/5, 7/5, 8/5;1/3125)))≈1.55968$

• Would you mind checking the numerical value for $n=6$ I gave in my so-called answer ? It looks that there will be aminimum ! I have probably an error somewhere. Thanks. – Claude Leibovici Jul 8 '18 at 7:13