# Math Help - Definite Infinite Integration

1. ## Definite Infinite Integration

I'm not sure if my title makes any sense... sorry, it has been a while since I've done cal.

I'm taking a Probability course and I'm reading one of my profs solutions that involves finding the moment-generating function of a continuous random variable. During the integration steps I got lost. Can someone please help understand how he gets from one step to another (I'll just include the part that confuses me):

$
\int^\infty_0 e^{ty}\alpha{e^{-\alpha y}} dy$

$
= \lim_{k\to\infty}\int^k_0 {\alpha}e^{\left({t-\alpha}\right)y}dy
$

$= \lim_{k\to\infty}[\frac{\alpha}{t-\alpha}e^{\left({t-\alpha}\right)y}]|(y=k,y=0)
$

Q1: why is it necessary to introduce $\lim_{k\to\infty}$ in the second line?

Q2: How does he get from the second line to the third? What method is that? Would it not be possible to simply take out all the constants from the integral and just have
$
= \lim_{k\to\infty}\frac{\alpha}{e^{\left({t-\alpha}\right)}}\int^k_0 e^{y}dy
$
?

Q3: Is there any quicker way to write the equations? It took me an hour to type this up using the code...

Thanks!

I'm not sure if my title makes any sense... sorry, it has been a while since I've done cal.

I'm taking a Probability course and I'm reading one of my profs solutions that involves finding the moment-generating function of a continuous random variable. During the integration steps I got lost. Can someone please help understand how he gets from one step to another (I'll just include the part that confuses me):

$
\int^\infty_0 e^{ty}\alpha{e^{-\alpha y}} dy$

$
= \lim_{k\to\infty}\int^k_0 {\alpha}e^{\left({t-\alpha}\right)y}dy
$

$= \lim_{k\to\infty}[\frac{\alpha}{t-\alpha}e^{\left({t-\alpha}\right)y}]|(y=k,y=0)
$

Q1: why is it necessary to introduce $\lim_{k\to\infty}$ in the second line?
It is notational convention. Whenever I do improper integrals with an infinite limit of integration I never do this, I just leave it as infinity. But it is up to your instructor to stipulate the notation he prefers.

Q2: How does he get from the second line to the third? What method is that? Would it not be possible to simply take out all the constants from the integral and just have
$
= \lim_{k\to\infty}\frac{\alpha}{e^{\left({t-\alpha}\right)}}\int^k_0 e^{y}dy
$
?
No, you are using a non-existant rule of exponents. I bet if you look closely again you will see why. And the method is just substitution.

Q3: Is there any quicker way to write the equations? It took me an hour to type this up using the code...
Nope, sorry. You will eventually learn and get faster. And just think, for all the time you put into actually typing the code out the more likely people will respond.

I'm not sure if my title makes any sense... sorry, it has been a while since I've done cal.

I'm taking a Probability course and I'm reading one of my profs solutions that involves finding the moment-generating function of a continuous random variable. During the integration steps I got lost. Can someone please help understand how he gets from one step to another (I'll just include the part that confuses me):

$
\int^\infty_0 e^{ty}\alpha{e^{-\alpha y}} dy$

$
= \lim_{k\to\infty}\int^k_0 {\alpha}e^{\left({t-\alpha}\right)y}dy
$

$= \lim_{k\to\infty}[\frac{\alpha}{t-\alpha}e^{\left({t-\alpha}\right)y}]|(y=k,y=0)
$

Q1: why is it necessary to introduce $\lim_{k\to\infty}$ in the second line?

[snip]
Because you cannot substitute $y = \infty$ into the anti-derivative since $\infty$ is not a number. You have to consider the limiting value of the anti-derivative as $y \rightarrow \infty$.

Contrary to the opinion of some, this is not a notational convention. It is a technical requirement. Although it's often omitted in arriving at the final answer, it is always implied and should always be understood.

I'm not sure if my title makes any sense... sorry, it has been a while since I've done cal.

I'm taking a Probability course and I'm reading one of my profs solutions that involves finding the moment-generating function of a continuous random variable. During the integration steps I got lost. Can someone please help understand how he gets from one step to another (I'll just include the part that confuses me):

$
\int^\infty_0 e^{ty}\alpha{e^{-\alpha y}} dy$

$
= \lim_{k\to\infty}\int^k_0 {\alpha}e^{\left({t-\alpha}\right)y}dy
$

$= \lim_{k\to\infty}[\frac{\alpha}{t-\alpha}e^{\left({t-\alpha}\right)y}]|(y=k,y=0)
$

[snip]

Q2: How does he get from the second line to the third? What method is that? [snip]
You have to remember how to integrate an exponential function.

$\int A e^{By} \, dy = \frac{A}{B} e^{By}$

I've omitted the arbitrary constant of integration since this result is being used to find a definite integral.

Warning: You're in for a world of pain unless you start extensively revising the calculus you've forgotten.