# How can we calculate the limit $\lim_{x \to +\infty} e^{-ax} \int_0^x e^{at}b(t)dt$?

I am looking at the following exercise:

Let the (linear) differential equation $y’+ay=b(x)$ where $a>0, b$ continuous on $[0,+\infty)$ and $\lim_{x \to +\infty} b(x)=l \in \mathbb{R}$.

Show that each solution of the differential equation goes to $\frac{l}{a}$ while $x \to +\infty$,

i.e. if $\phi$ is any solution of the differential equation, show that $\lim_{x \to +\infty} \phi(x)=\frac{l}{a}$.

That’s what I have tried:

The solution of the differential equation will be of the form:

$\phi(x)=ce^{-ax}+e^{-ax} \int_0^x e^{at}b(t) dt$

$\lim_{x \to +\infty} c e^{-ax}=0$

So, $\lim_{x \to +\infty} \phi(x)=\lim_{x \to +\infty} e^{-ax} \int_0^x e^{at}b(t)dt$

How can we calculate the limit $\lim_{x \to +\infty} e^{-ax} \int_0^x e^{at}b(t)dt$ ?

EDIT:
$$\lim_{x \to +\infty} b(x)= l \in \mathbb{R}$$

That means that $\forall \epsilon>0$, $\exists n_0 \in \mathbb{N}$ such that $\forall n \geq n_0$: $|b(x)-l|< \epsilon \Rightarrow – \epsilon< b(x)-l< \epsilon \Rightarrow l- \epsilon<b(x)<l+ \epsilon$.

We pick $\epsilon=\frac{l}{2}$ and so we have that: $b(x)>\frac{l}{2}$.

Thus $\int_0^x e^{at} b(t)dt> \int_0^x e^{at} \frac{l}{2} dt=\frac{l}{2} \int_0^x e^{at}dt$

• If $l>0$ then $\frac{l}{2} \int_0^x e^{at}dt \to +\infty$ and so $\int_0^x e^{at} b(t)dt$ diverges.
• If $l<0$ then $\frac{l}{2} \int_0^x e^{at}dt \to -\infty$ and so $\int_0^x e^{at} b(t)dt$ diverges.

In both of the above cases, we can use the Fundamental Theorem Of Calculus:

$$\lim_{x \to +\infty} e^{-ax} \int_0^x e^{at} b(t) dt= \lim_{x \to +\infty} \frac{\int_0^x e^{at} b(t) dt}{e^{ax}}=\lim_{x \to +\infty} \frac{e^{ax} b(x)}{a e^{ax}} \to \frac{l}{a}$$

What can we say for the case $l=0$?

#### Solutions Collecting From Web of "How can we calculate the limit $\lim_{x \to +\infty} e^{-ax} \int_0^x e^{at}b(t)dt$?"

One idea is to break up $\int_0^x e^{at}b(t)dt$ into two parts,

$$\int_0^M e^{at}b(t) \ dt + \int_M^x e^{at}b(t) \ dt$$

where $M$ comes from the limit of $b$. Namely, given $\epsilon > 0$, there exists an $M$ such that

$$x > M \ \Rightarrow |b(x) – L| < \epsilon$$

I.e., $x > M$ implies $L – \epsilon < b(x) < L + \epsilon$. This enables you to put upper and lower bounds on the second integral.

The first integral when multiplied by $e^{-ax}$ goes to zero.

Although I just saw Git Gud’s hint to use L’Hôpital. That would be simplier 😉

L’Hôpital’s rule can be applied if it is only known that the limit of the denominator is $\infty$. Then

$$\lim_{x \to +\infty} e^{-ax} \int_0^x e^{at}\,b(t)\,dt=\lim_{x\to+\infty} \frac{\int_0^x e^{at}b(t)dt}{e^{ax}}=\lim_{x\to+\infty} \frac{e^{ax}\,b(x)}{a\,e^{ax}}=\frac{l}{a}.$$