Recall the definition of the Gamma function:
Using integration by parts, prove the functional equation:
Then use mathematical induction to prove that for positive integers we have
Incomplete.
Recall the definition of the Gamma function:
Using integration by parts, prove the functional equation:
Then use mathematical induction to prove that for positive integers we have
Incomplete.
Test the following series for convergence or divergence. Justify the decision.
First, we make the substitution , which gives us . Then we have
This integral we evaluate using integration by parts with
Therefore, we have
So, for the definite integral from to we have
But then the series is a telescoping series with
Hence,
Hence, the series converges.
Prove the following integral formulas.
This gives us
This gives us
This gives us
Proof. The proof is by induction. We have already established the case (and and ). Assume then that the formula holds for some positive integer . We then consider the integral
Integrating by parts, we let
Therefore, integrating by parts and using the induction hypothesis we have,
Therefore, the formula holds for , and hence, for all positive integers
Let
We compute,
Then, since
we have
and
Hence, the set of such that is all real .
Then, using the translation invariance of the integral,
Proof. Making the substitution , we have
Proof. To prove this identity we integrate by parts, letting
Therefore,
Proof. For the final identity, we use the substitution , . This gives us
Now, we use integration by parts with
This gives us
Try to use integration by parts to evaluate the integral
If we let
Then we have
If we try to integrate by parts again with and then we’ll end up with another integral, this time with . That will continue, so we’ll just keep getting integrals that we can’t evaluate.
On the other hand, if we try letting
Then we have
Continuing along that route, we’ll keep getting integrals of where keeps getting larger.
In both cases, we keep getting increasing complicated integrals that we can never evaluate.
Compute the following integral.
First, we want to make the substitution
Therefore, we have
Now, we use integration by parts, letting
Therefore, we have
Moving the back to the left side we have
Plugging this back in above we have,
Then we note that
(If you don’t remember your trigonometry, which I usually don’t, you can figure things like this out by drawing the triangle. Since means you have a right triangle with legs of length and . That means the hypotenuse has length . So, which then gives the above result for .)
Plugging this back in we then have
(Where we absorbed into the constant in the final line, since it is just a constant.)
Establish that the following integral formula is correct:
Proof. We can evaluate this integral using integration by parts. Let
Then we have
Next, we must evaluate the integral on the right. For this integral we have
Now, we make a substitution, let , . Therefore,
Combining this with the equation above we then have
Establish that the following integral formula is correct:
Proof. We’re going to integrate by parts twice to evaluate this integral. First,
So, we have
Now, we evaluate the integral on the right using integration by parts again. Let,
Therefore,
So, putting this back into the equation above we have
Establish that the following integration formula is correct:
Proof. We integrate by parts, letting
(Where we established the formula for the derivative of in this exercise, Section 6.22, Exercise #5). Therefore,
In the previous exercise we showed that
Therefore,
Establish that the following integration formula is correct:
Proof. First, we want to integrate by parts. Let
Therefore, we have
Now, to evaluate this integral we break it into pieces (to get rid of the ). Since we have
(Since is only defined for we don’t need to worry about the cases .) We then evaluate the two pieces of the integral separately. If ,
Now, we note that
Therefore, we have an integral of the form and so,
So, for we have
For the case everything is identical except we have a negative sign (since has a negative sign when ) so for we have
Therefore (since if and if ) we have