Consider the intergal $I=\int_{1}^{\infty}e^{ax^2+bx+c}dx$, where $a,b,c$ are constants. When does the integral converge?

Question

Consider the integral $I=\displaystyle\int_{1}^{\infty}e^{ax^2+bx+c}dx$, where $a,b,c$ are constants. When does the integral converge?

As usual, these are alien concepts to me, it gets tough to understand, but I realize, if the integral value is finite, we will have a solution. But, I cannot do anything more.

Answer 1

The exponent of the exponential is $ax^2+bx+c=a(x+b/2a)^2-(b^2-4ac)/4a$ for $a\ne 0$. Thus, the integral becomes

$$\int_1^{\infty} e^{ax^2+bx+c}dx=e^{-(b^2-4ac)/4a}\int_1^{\infty} e^{a(x+b/2a)^2}dx=e^{-(b^2-4ac)/4a}\int_{1+b/2a}^{\infty} e^{ax^2}dx$$

Thus, the integral converges (diverges) for $a<0$ ($a>0$) for all $b$ and $c$.

For $a=0$, the exponent of the exponential is $bx+c$. Thus, the integral becomes

$$\int_1^{\infty} e^{bx+c}dx=e^{c}\int_1^{\infty} e^{bx}dx$$

and the integral converges (diverges) for $b<0$ ($b\ge 0$).

Answer 2

A function that tends to $\infty$ as $x\to\infty$ cannot be integrable over $(1,\infty)$. Therefore either $a<0$, or $a=0$ and $b<0$, or $a=b=0$, as these are the only ways to make the function bounded as $x\to\infty$.

$a=b=0$ leaves you with a constant function, clearly not integrable over $(1,\infty)$.

$a=0$ and $b<0$ leaves you with an integrable function over $(1,\infty)$, seen by just writing an antiderivative and seeing its limit as $x\to\infty$.

$a<0$ leaves you with an integrable function over $(1,\infty)$, because for large enough $x$ you have a positive function even smaller than the $a=0$ and $b<0$ case.