# (another) Challenging improper integral [closed]

$$\int_0^\infty y^{1/2}e^{-y^3}\,dy$$ It is in the section with the gamma function if that helps. Thanks!

Let $y = t^{2/3}$. Then $dy = (2/3)t^{-1/3}\, dt$, and hence

$$\int_0^\infty y^{1/2}e^{-y^3}\, dy = \int_0^\infty t^{1/3}e^{-t^2} \cdot \frac{2}{3}t^{-1/3}\, dt = \frac{2}{3}\int_0^\infty e^{-t^2}\, dt = \frac{2}{3}\cdot \frac{\sqrt{\pi}}{2} = \frac{\sqrt{\pi}}{3}.$$

• Kobe. Nice fortuitous sub! Changes from a Gamma function to an Error function integral. – Mark Viola Mar 5 '15 at 0:43

$$\int_0^\infty y^{1/2}e^{-y^3}\,dy = \int_0^\infty y^{1/6} e^{-y}(2/3 y^{-2/3})\,dy = 2/3 \int_0^\infty y^{-1/2}e^{-y}\, dy = 2/3\Gamma(1/2).$$

\begin{split} \int_0^{\infty}y^{1/2}e^{-y^3}\text{d}y&=\frac{2}{3}\int_0^{\infty}e^{-y^3}\text{d}y^{3/2}\\ &=\frac{2}{3}\int_0^{\infty}e^{-t^2}\text{d}t\\ &=\frac{2}{3}\sqrt{\int_0^{\infty}e^{-t^2}\text{d}t\times\int_0^{\infty}e^{-m^2}\text{d}m}\\ &=\frac{2}{3}\sqrt{\iint_{[0,\infty]\times[0,\infty]}e^{-(t^2+m^2)}\text{d}t\text{d}m}\\ &=\frac{2}{3}\sqrt{\int_{0}^{\infty}\int_0^{\pi/2}e^{-r^2}r\text{d}r\text{d}\theta}\\ &=\frac{2}{3}\sqrt{\frac{\pi}{4}\int_{0}^{\infty}e^{-r^2}\text{d}r^2}\\ &=\frac{\sqrt{\pi}}{3} \end{split} Hope this can help you.

After the substitution $x = y^3$, $$\int_0^\infty y^{1/2}e^{-y^3}dy = \frac{1}{3}\int_0^\infty x^{-1/2}e^{-x}dx = \frac{\Gamma(1/2)}{3} = \frac{\sqrt{\pi}}{3}.$$

what happens if we make the change of variable $$u = y^3, y = u^{1/3}, dy = \frac 13 u^{-2/3} \, du$$ so that $\int_0^\infty y^{1/2}e^{-y^3}\, dy$ is transformed into $$\int_0^\infty u^{1/6}e^{-u} \frac 13 u^{-2/3} \, du = \frac 13 \int_0^\infty u^{-1/2}e^{-u} \, du = \frac 13 \sqrt \pi$$

• @Dr.MV, i checked. i think i have it right. – abel Mar 5 '15 at 0:48
• I had seen $sqrt(\pi/2)$ when I wrote the comment. It is correct now. But I congratulated Kobe for a less obvious substitution that augments the integral type from Gamma to Gaussian. – Mark Viola Mar 5 '15 at 0:51