You are Here: Home >< Maths

# The Proof is Trivial! Watch

1. (Original post by Jkn)
Problem 269***

Using the previous problem as a hint,

Find functions f and g such that
Dare I give:
, ?
Apologies if I've misunderstood - I flee from contour integration at every possible opportunity (not that there have been many such opportunities - I really mean that any time it might have been useful to learn contour integration, I ran away from it like the wind).
2. (Original post by Smaug123)
Is this possible without contour integration? Contour integration is horrible
Hmm, I'd imagine it would be doable with Taylor's series? Not too sure.
(Original post by Smaug123)
Dare I give:
, ?
Apologies if I've misunderstood - I flee from contour integration at every possible opportunity (not that there have been many such opportunities - I really mean that any time it might have been useful to learn contour integration, I ran away from it like the wind).
I've modified the question now - the point of the question being to express the mysterious Zeta function in terms of less-mysterious 'elementary' functions which can, in turn, be used to approximate () by using integral approximations

I learnt it this morning but I'm still feeling a bit shaky (hoping people will engage in discussion on ASOM with me with regards to a few problems!) I'm probably not quite at the point where I could have done the problem LOTF just solved, though I did think it worth posting as I knew a solution would not take long to emerge.
3. Problem 270

what is: ?
4. (Original post by Hasufel)
Problem 270

what is: ?
Solution 270

Now, note that

5. (Original post by Felix Felicis)
x
Basically the same as yours but more direct:

Solution 270 (2)

Let ,

Spoiler:
Show
Whilst the substitution may seem arbitrary, it is best spotted by seekng a substitution that has that property that it combines powers of x, the derivative and . The first two will combine nicely regardless but the third is more difficult to find (we must seek the inverse). Our substation works nicely as .
We can generalise by using the same substitution,

6. which is known as the Euler-Mascheroni constant.

Problem 271
***

Prove that .

Note that I require you to prove any non-* theorems/results that you use in your proof.

Problem 272*

Prove that

Problem 273***

Evaluate

Prove all non-* theorems, as before.

Problem 274***

Evaluate

Problem 275***

Evaluate
Problem 165***

Evaluate .
Not sure this is right as my final answer looks messy, so let me know.

Solution 165

Let :

Comparing the integrals to the Beta function, using the Gamma function representation of the Beta function as well as Euler's reflection formula, we get:

Let and then :

Splitting the denominator into its complex routes and integrating:

Simplifying the result as much as we can and combining and , we get:

8. (Original post by Jkn)
which is known as the Euler-Mascheroni constant.

Problem 271
***

Prove that .

Note that I require you to prove any non-* theorems/results that you use in your proof.
Does that include proof of differentiation under the integral sign?
9. (Original post by Felix Felicis)
Does that include proof of differentiation under the integral sign?
Sorry, no! I suppose I meant non-** theorems. Essentially I am asserting that is not quotable
10. Sort of a mixture between the last two solutions but I thought the route it leads down seems different enough to warrant a post:

Solution 168
(3)

Beginning as in solution (1), we can deduce that

Let

We can generalise the last integral in a rather interesting way:

We now note the Taylor's series of about
for where denotes the nth Euler number.

By considering the coefficients of this series, we can find closed forms for the above integral:

for all .

Note that this implies that the above integral is equal to 0 whenever n is odd (this is implied both by the Taylor's series and the fact that all nth Euler numbers are 0 when n is odd).

So, for example,

Corollary: Letting n=0 gives us yet another way to deduce that
11. Solution 274

Let

Now, consider the gamma function:

which evaluated at gives

We have that

12. (Original post by Felix Felicis)
x
Nice stuff man, don't you just love gamma functions, DUTIS and all that crap?

Btw, do you like my two monster posts?
13. (Original post by Jkn)
Nice stuff man, don't you just love gamma functions, DUTIS and all that crap?

Btw, do you like my two monster posts?
Gotta love it man, it's some good stuff

I do! Very nice I especially liked the extension to 168 And while I'm at it, I was looking through some old stuff and saw your method on analysing the Gaussian integral, that was quite nice too
14. (Original post by Felix Felicis)
Gotta love it man, it's some good stuff

I do! Very nice I especially liked the extension to 168 And while I'm at it, I was looking through some old stuff and saw your method on analysing the Gaussian integral, that was quite nice too
Ikr! You're going to absolutely destroy at uni next year. Are you applying (or should I saying going) to Cambridge and, if so, which college?

Thanks man! I've got into the habit of extending everything now, it's so satisfying! I've fallen in love with Euler's reflection formula btw! I was messing about with the problem as I was trying to work through Mladenov's contour integration method, and then I though I would ass that to the mix. I then thought that I probably looked like a bit of a **** as I was bringing nothing new to the table and then realised that, whist repeated differentiation tends to be tedious, the Taylor's series was an awesome shortcut! We don't use Taylor's (or Laurent) series enough on this thread, perhaps you could dig up some fun questions?

Cheers but I felt like a bit of an idiot right after I finished as I realised that it was entirely unnecessary given the fact that all I had to do was use a trigonometric substitution to trivially evaluate the appropriate value of the Beta function (though I left it there as the digamma method of evaluating that integral was interesting in itself ). Incidentally, it was that precise integral that was the entirety of STEP II Q2 2013 (which makes it really satisfying that I found it of my own accord)

I'm looking forward to being taught how to evaluate the Gaussian next year I'll stroll up like "yeah I found like 10 ways to do this ever the summer "

I hope we are not being too narrow by being obsessed with integration and special functions..

Oh, btw, if you liked some of those posts you might like the one I wrote about the E-M constant (etc..) a few pages back on ASOM
15. Problem 276

(simple one, really)

if which choice of both and make as large as posilble?
16. (Original post by Hasufel)
Problem 276

(simple one, really)

if which choice of both and make as large as posilble?
Two numbers which add to make 50 can be written in the form (25+k),(25-k) so that their product is:

Clearly for this to be maximised k=0 (since a square is either positive or 0). Hence the largest product is 625.
17. (Original post by Hasufel)
Problem 276

(simple one, really)

if which choice of both and make as large as posilble?
Solution 276:
I say:
where n is extraordinarily large.
As:

yet:
18. (Original post by Hasufel)
Problem 276

(simple one, really)

if which choice of both and make as large as posilble?
Am I being silly? Can't you just change into , then just maximize this to get ?
19. (Original post by joostan)
I say:
where n is extraordinarily large.
As:

yet:
I prefer to keep it real .
20. Problem 276 *

Evaluate

Spoiler:
Show
I found this in a Hong Kong A-level paper but they walked you through to a solution. If you want a big hint check out the spoiler.
Spoiler:
Show
Consider a substitution of the form for some suitable

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: December 11, 2017
Today on TSR

### Is this person a genius?

...with these A Level results?

### I think I'm transgender AMA

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.