# Thread: Find all values of a for which the following improper converges.

1. ## Find all values of a for which the following improper converges.

$\Large I = \int_{0}^{\infty} e^{ax} cos(x) dx$
I calculated the indefinite integral and founded it equals $\large \dfrac{e^{ax}}{a^{2}+x} (sin(x)+acos(x) ) + C$
Now, replacing $\infty$ by t and evaluate the limit I faced this limit :
$\large \dfrac{1}{a^{2}+1} \lim_{t \to \infty} e^{at} \cdot (sin(t)+a cos(t)) - a )$
I stopped at this limit.

2. ## Re: Find all values of a for which the following improper converges.

That will definitely diverge unless a is negative.

3. ## Re: Find all values of a for which the following improper converges.

Why is that?
ok if a is positive then e^at will be infinity, but inside brackets will be not defined as sin and cos are not exist at infinity.
If a is negative it will be like
$\large \dfrac{sin(t)+acos(t)-a}{e^{kt}}$ with k=-a>0
How to calculate this limit algebraically?

4. ## Re: Find all values of a for which the following improper converges.

Originally Posted by TWiX
Why is that?
ok if a is positive then e^at will be infinity, but inside brackets will be not defined as sin and cos are not exist at infinity.
If a is negative it will be like
$\large \dfrac{sin(t)+acos(t)-a}{e^{kt}}$ with k=-a>0
How to calculate this limit algebraically?
\begin {align*} &\displaystyle \int_0^\infty ~e^{a x}\cos(x)~dx = \\ \\ &\left(\lim \limits_{x \to \infty} \dfrac{e^{a x}(a \cos(x)+\sin(x))}{1+a^2}\right) - \dfrac{e^{0}(a \cos(0)+\sin(0))}{1+a^2} = \\ \\ &\left(\lim \limits_{x \to \infty} \dfrac{e^{a x}(a \cos(x)+\sin(x))}{1+a^2}\right) -\dfrac{a}{1+a^2} \end{align*}

if $a>0$ it should be pretty clear that the limit on the left hand side blows up.

if $a=0$ this is the limit of a periodic function which thus doesn't exist

if $a<0$ then this is a damped exponential which converges to 0 and thus

$a < 0 \Rightarrow \displaystyle \int_0^\infty ~e^{a x}\cos(x)~dx = -\dfrac{a}{1+a^2}$

5. ## Re: Find all values of a for which the following improper converges.

What do you mean mathematically when a>0 the limit will "blow up"?
Give me mathematical explaination here.
for x-->infinity e^ax is infinity but we have cos(x) and sin(x) there.

6. ## Re: Find all values of a for which the following improper converges.

Originally Posted by TWiX
What do you mean mathematically when a>0 the limit will "blow up"?
Give me mathematical explaination here.
for x-->infinity e^ax is infinity but we have cos(x) and sin(x) there.
$\lim \limits_{x\to 0}~e^{a x} = \begin{cases} \infty &a > 0 \\ 1 &a=0 \\ 0 &a<0 \end{cases}$

while it's true that the trig terms cause the value of the expression to oscillate it's still true that

$a>0 \Rightarrow \forall \epsilon > 0,~\exists x \ni e^{ax}(a \cos(x) + \sin(x)) > \epsilon$

i.e. the expression "blows up"