this is a solution to my all-in-one calculus question and let's get started with this
integral right here as we can see we have a square root of T here and also a negative
there let's just go ahead and do a u sub and see what happens and of course that's the U equal to
T no that doesn't do us anything right so try square root of T and of course Square
both sides Square t equals U squared and then differentiating both sets we get DT equals 2 u
d u and now we can take this integral from the T world to the U world when T is equal to zero
put it here that means you will be starting at zero and then when we put Infinity to the
square root of T we get infinity so U will be Infinity right here square root of T is
our U and then we have e to the negative T is U squared and the last part is DT which is 2 u d u
so we multiply the U and U together no don't do that because otherwise we are doing a chance to
use integration by parts because this power can integrate and of course just go ahead and
differentiate this so let's see what happens so I'm going to just put down D and I for the DI
method and the Plus for minus let's differentiate to you and then we'll integrate U times e to the
negative U squared differentiating this will get you integrating this to another substitution let
W equal to negative U squared then we get negative 2 u d u this is only at U so let's multiply this
by negative two and of course divided by that right away so that the top is just going to
give us e to the negative U squared and you can double check if you differentiate this you do end
up with setup and of course don't forget the negative one half the consider multiple well
after we have this we can construct our insert multiply the diagonal that would be the first
part of the answer and I see we will get two times that will just cancel and we will have a negative
U times that and let's just go ahead and put this on the bottom so we have e to the U squared
this part does not have any integral anymore and we will just have to evaluate it from 0 to
infinity and then we are going to multiply this row and put that inside of an integral
negative times a negative gives us a positive and an integral goes from 0
to Infinity 2 times 1 half cancel and we just have e to the negative U squared in the U world
now let's take a look right here poor Infinity to U here and U here technically we have to take the
limit and if we do that the limit as U goes to Infinity of negative U over e to the U squared
you can either use the list and argue that exponential function like this each the U
squared is so much bigger than negative U and you end up with 0 or you can just use lapitos
rule differentiate the top you get negative one differentiate this you get 2u times e to the U
squared and then U equals to Infinity used to end up with zero so that will give us 0 and when we
put 0 into here thanks to the U on the top we will also get zero so the first part is just equal to
zero so nice and then for this part it's just as nice the integral goes from 0 to Infinity of this
e to the negative U squared we have done that recently huh yes so let me just make that note
right here for you guys the version that we did was negative Infinity to past infinity
and in my previous video I use x so I will use x right here as well negative x squared
this right here if we go all the way we end up with square root of Pi this
is the famous gaussian integral in that video I use the LaPlace methods to do it and it did
not involve any polar coordinates so go ahead and check that if you haven't answered already
however this right here is only halfway but because we go from 0 to infinity and since this
right here converges we can just go ahead and divide this by two and we get square root of pi
over two and of course all together we just get square root of pi over 2.
this means X is just approaching square root of pi over 2. and you might be wondering why
didn't we say that X is approaching square root of pi over 2. well I just want to show off this
integral all right next let's look at inside here we have the power series you can try to figure out
what the function is right now but don't do that because we are differentiating this power series
twice and this is actually one of the best things in calculus I can differentiate any power series
all day let's see how this works differentiate this one time just look at X to hyper power use
the power rule bring the power to the front and we get 4 um plus four and of course minus
one that's differentiating power series and you see one plus four cancel out so nice huh
and let's do it again because that's the second derivative so right here you see
the result is 4 M plus three we overtake this and then putting the front again and then minus
one and we get one plus three here and this is not cancel and the final power here is for um plus two
so this part right here as we can see it's going to give us what
still a power series and then it goes from zero to Infinity but we only have the two one plus one on
the bottom and let's write it down like this two and plus one and then the top is one and let's put
the X to this power on the side four and plus two another question is what is this function
well let's see we have an odd number here huh 2 and plus one but it does not have
the factorial so don't think about sign War hyperbolism don't do that
this right here it's not even alternating if we have a negative 1 to the nth power and then
over two n plus one then you can kind of suspect we may have like in the the sine function but no
so let's see what this is though when you do not have the factorial you should think
about the inverse tangent but this is not alternating so what is this
don't worry this is just the inverse tangents person named the inverse hyperbolic tangent
but the powers they don't match this is 4M plus two this is two and plus one
don't worry check this out this right here we can write this on purpose like n goes from 0
to infinity and this right here is one over two M plus one for the X let's put yes x square and
then this right here becomes 2m plus one yes and we can say that this is not regular tangent but
relatively hyperbolic tangent and this lesson set is the inverse version and the input is x square
the interval from negative one to one two nine including points though this right here is the
interval convergence for this power series meaning that we can put any x value between negative one
and one to this and it will work square root of pi over 2 is positive and it's also less than one so
it's certainly on this interval therefore ladies and gentlemen our final answer is we just put this
in here so all that is just equal to the inverse hyperbolic tangent of square root of pi over 2
and then we square that AKA inverse hyperbolic tangent of pi over four yay can we simplify
this a little bit you can write the natural log version of the inverse hyperbolic tangent but
I think that's I'll just leave the answer like that this is approximately one something though