# How to evaluate this definite integral and find limit when upper limit of definite integral tends to infinity

Find $$\lim_{x\to\infty} x^2\int_0^x e^{t^3-x^3}\,dt$$

A. $1/3$
B. $2$
C. $\infty$
D. $2/3$

I tried to integrate the definite integral first directly in hope of putting limits afterwards but failed. I used integration by parts twice after taking substitutipn of entire term in e power but in the end everything cancelled and I got 0 = 0 . Where am I wrong ? May someone help?

• can you use $$\LaTeX$$ please? – Dr. Sonnhard Graubner Nov 16 '16 at 12:21

You can rewrite the limit as $$\lim_{x\to\infty}\frac{\displaystyle \int_0^x e^{t^3}\,dt}{e^{x^3}/x^2}$$ and apply l'Hôpital, because the denominator has limit $\infty$; the derivative of the denominator is $$\frac{3x^4e^{x^3}-2xe^{x^3}}{x^4}=\frac{3x^3-2}{x^3}e^{x^3}$$ so, by applying the fundamental theorem of calculus, you have to compute $$\lim_{x\to\infty}\frac{e^{x^3}}{\dfrac{3x^3-2}{x^3}e^{x^3}}$$
• @RaghavSingal You don't need to, but it is easy nonetheless: the function is positive and unbounded. Also $\lim_{z\to\infty}e^z/z=\infty$, so also the denominator tends to infinity. – egreg Nov 16 '16 at 13:30
Here's a more scenic route $$\lim\limits_{x\to\infty}x^2\int_0^x e^{t^3-x^3}\ \mathrm dt$$ $$=\lim\limits_{x\to\infty} \frac{x^2}{e^{x^3}}\int_0^x e^{t^3}\ \mathrm dt$$ $$=\lim\limits_{x\to\infty}\frac{x^2}{e^{x^3}}\int_0^x \sum\limits_{k=0}^{\infty}\frac{\left(t^3\right)^k}{k!}\ \mathrm dt$$ $$=\lim\limits_{x\to\infty}\frac{x^2}{e^{x^3}}\sum\limits_{k=0}^{\infty}\frac{1}{k!}\int_0^x t^{3k}\ \mathrm dt$$ $$=\lim\limits_{x\to\infty}\frac{x^2}{e^{x^3}}\sum\limits_{k=0}^{\infty}\frac{x^{3k+1}}{k!(3k+1)}$$ $$=\frac13\lim\limits_{x\to\infty}\frac{x^3}{e^{x^3}}\sum\limits_{k=0}^{\infty}\frac{\left(x^3\right)^k}{k!\left(k+\frac13\right)}$$ $$=\frac13\lim\limits_{z\to-\infty}\frac{\left(-z\right)}{e^{-z}}\sum\limits_{k=0}^{\infty}\frac{\left(-z\right)^k}{k!\left(k+\frac13\right)}$$ $$=\frac13\lim\limits_{z\to-\infty}\frac{\left(-z\right)}{e^{-z}}z^{-\frac13}\gamma\left(\frac13, z\right)$$ $$=\frac13\lim\limits_{z\to-\infty}\frac{-\gamma\left(\frac13, z\right)}{e^{-z}z^{\frac13-1}}$$ $$=\frac13\lim\limits_{z\to-\infty}\frac{\Gamma\left(\frac13, z\right)-\Gamma\left(\frac13\right)}{e^{-z}z^{\frac13-1}}$$ $$=\frac13\left(1-0\right)=\frac13$$