The hard integral thread.

Original post by atsruser

I posted this (with slightly different limits) on the Proof is Trivial thread where it was left unsolved, so I'll repeat it here. It's well within the capabilities of an A level student:

Give a geometrical argument to show that, if

x^2+y^2=1

, then

\displaystyle \int_{\frac{\sqrt{2-\sqrt{2}}}{2}}^{ \frac{\sqrt{2+\sqrt{2}}}{2} } \frac{1}{\sqrt{1-x^2}} \ dx = \frac{\pi}{4}

You'll probably need to draw a diagram to show the geometrical bit.

Time to clear this one up, I think:

Scan_20160120 (2)-crop.png

We are interested in the quantity

\frac{dx}{\sqrt{1-x^2}} = \frac{dx}{y}

as we want to integrate it between the two values of

x

supplied in the limits.

In picture (1) I've drawn a picture showing the relationship between the infinitesimal

dx, dy, ds

, with

ds

being the change in arc length for this curve. Consider triangles OAC and BDC. Look at picture (2) for a zoomed copy of BDC.

We can see that as

dx \to 0

we must have that CD

\to ds

and that thus angle OCD

\to \pi/2

. From triangle OAC, we have

\alpha + \theta = \pi/2

, so we must thus have, in the limit, angle BCD

\to \alpha

and also angle BDC

\to \theta

.

Thus triangles OAC and BDC are "ultimately similar", and we have that:

\frac{ds}{dx} = \frac{1}{y} \Rightarrow ds = \frac{dx}{y}

But note that for the unit circle

ds = 1 d\alpha = d\alpha

- this tells us that the integrand, when viewed from an infinitesimal point-of-view, simply measures the change in angle caused by a change in

x

.

We also have

\cos \alpha = x \Rightarrow \alpha = \cos^{-1} x

and in addition note that as

x

increases,

\alpha

decreases, so putting this all together, we have:

\displaystyle \int_{\frac{\sqrt{2-\sqrt{2}}}{2}}^{ \frac{\sqrt{2+\sqrt{2}}}{2} } \frac{1}{\sqrt{1-x^2}} \ dx = \int_{\alpha_1}^{\alpha_2} -d\alpha = \alpha_1-\alpha_2

The only mystery left is finding the two angles, if you don't recognise them from their cosines. But we can do that by writing:

\cos \alpha_2 = \frac{\sqrt{2+\sqrt{2}} }{2} \Rightarrow 2\cos^2 \alpha_2 -1 = \cos 2\alpha_2 = \frac{\sqrt{2}}{2} \Rightarrow \alpha_2= \frac{\pi}{8}

and

\cos \alpha_1 = \frac{\sqrt{2-\sqrt{2}} }{2} \Rightarrow 2\cos^2 \alpha_1 -1 = \cos 2\alpha_1 = -\frac{\sqrt{2}}{2} \Rightarrow \alpha= \frac{3\pi}{8}

So we have the integral equals

\frac{3\pi}{8}-\frac{\pi}{8} = \frac{\pi}{4}

by geometry (or by infinitesimal geometry, at least).

Reply 783

Star-girl

12

Original post by atsruser

Hmm. Bizarrely over-aggressive responsive. I do apologise. There I was, thinking that I was giving a vaguely useful suggestion to help someone improve their integration, when really I was being an evil, bastard, patronising ... bastard!!!!.

Don't worry - I shall delete the offending post, and shall take the greatest of care not to patronise you in the future, by the simple expedient of ignoring your future posts.

Total overreaction to being called out and I said never said you were being an "evil bastard" so don't know where you got that one from.

Whatever you obviously don't care anyway with you sarcastic fake response, so... bye.

Reply 784

Original post by Kummer

That's slightly easier than mine as well as you deal with no partial fractions. I couldn't agree more with your advice, btw. Orthogonality helped me tame this beast earlier in this thread.

You may wish to look at http://www.thestudentroom.co.uk/showpost.php?p=60397285&postcount=3399 for another similar one that I posted elsewhere. (I put up a solution later though, if you can't be bothered with it).

And thinking of orthogonality, I guess there should be plenty of scope for more hard integrals with other orthogonal sets - I really need a good book that covers all of the standard special functions though - I don't have one. Any suggestions?

Reply 785

Indeterminate

3

Evaluate

\displaystyle \int_{0}^{\infty} \dfrac{1}{x+1} \arctan \left(\dfrac{2\pi}{x - \ln(x) + \ln(\pi/2)}\right) \ dx

Reply 786

Why does almost every integral in this thread has limits?

Reply 787

shamika

16

Original post by atsruser

You may wish to look at http://www.thestudentroom.co.uk/showpost.php?p=60397285&postcount=3399 for another similar one that I posted elsewhere. (I put up a solution later though, if you can't be bothered with it).

And thinking of orthogonality, I guess there should be plenty of scope for more hard integrals with other orthogonal sets - I really need a good book that covers all of the standard special functions though - I don't have one. Any suggestions?

How about a classic?

https://ia800504.us.archive.org/5/items/HandbookOfMathematicalFunctions/AMS55.pdf

Reply 788

trapking

20

Original post by TeeEm

Why does almost every integral in this thread has limits?

i was asking myself the same thing!

Spoiler

Reply 789

Original post by trapking

i was asking myself the same thing!

Spoiler

We have to wait for some plausible explanations

Reply 790

trapking

20

Original post by TeeEm

We have to wait for some plausible explanations

Indeed, some of this looks like chinese to me :lol:

Reply 791

Zacken

22

Original post by TeeEm

Why does almost every integral in this thread has limits?

Probably a shot in the dark here, but most of these integrals require some crazy substitution which make the indefinite integral look really ugly, however with limits in, they simplify to nice numbers.

Also with improper limits like 0 to infinity, that gets rid of some "clutter" that makes the end result aesthetically pleasing, then again... what do I know. :tongue:

Reply 792

ThatPerson

18

Original post by TeeEm

Why does almost every integral in this thread has limits?

Because integrating a function where an antiderivative exists in terms of elementary functions would be too easy :tongue:

.

Reply 793

Original post by Zacken

Probably a shot in the dark here, but most of these integrals require some crazy substitution which make the indefinite integral look really ugly, however with limits in, they simplify to nice numbers.

Also with improper limits like 0 to infinity, that gets rid of some "clutter" that makes the end result aesthetically pleasing, then again... what do I know. :tongue:

It is kind of a reason.

Reply 794

shamika

16

Original post by atsruser

I think that you're being a bit pessimistic there. It's true that it couldn't be given simply in the form "Do this integral: ...". But it could be a nice multi-part question, with part 1 being "sum this complex series", part 1 being some hints about the orthogonality of sin/cos, and the third part being the integral itself. That would be a decent STEP III question, I think.

Hmmm...

I find that a lot of the questions on the thread are doable (for me) precisely because I've gone through a maths degree - looking for orthogonlity of sin/cos is obvious if you've done any Fourier transforms (and even more so if you took a course in Fourier analysis). Even when I do STEP, I tend to first think of undergrad sledgehammers to crack the nut (of course, I then go and do it the intended way).

What you're saying is right of course - you could figure it out without knowing anything about Fourier stuff, but it's definitely a lot harder without being trained to think that way.

I guess what I'm trying to say is that helping out with STEP prep for the last few years, I've learned that a degree just broadens how people think of a problem, even for the most talented of mathmos. With the additional parts, it's a lot more reasonable of course.

Anyway, I've just waffled a lot. I'm going to look at the special functions book to put me asleep, I'm tired! :tongue:

Reply 795

Original post by ThatPerson

Because integrating a function where an antiderivative exists in terms of elementary functions would be too easy :tongue:

.

In my humble opinion it is the opposite.

Indefinite integration is a Caveman which kills you brutally ...
Definite integration is is humane lethal injection.

Reply 796

Lord of the Flies

Original post by TeeEm

Why does almost every integral in this thread has limits?

Because antiderivatives are mechanical, and therefore of little interest. A computer program can figure out antiderivatives, if they exist - they cannot however, do some of the things done on this thread.

Would you really be amazed if I gave you the antiderivative of

\sqrt[2015]{\tan x}?

Highly doubt it. Doable, and a pain in the arse for sure, but not actually worth anything. Pages of meaningless drivel.

If however, I told you that

x^{x}(1-x)^{1-x}\sin \pi x

evaluates to

\pi e \cdot 24^{-1}

over

(0,1)

, then I am sure you - or anyone else here - would think "oh cool, might try and prove that".

Reply 797

Original post by Lord of the Flies

... then I am sure you - or anyone else here - would think "oh cool, might try and prove that".

Wow, I have been put back into my place...

Reply 798

Original post by shamika

What you're saying is right of course - you could figure it out without knowing anything about Fourier stuff, but it's definitely a lot harder without being trained to think that way.

Well, I'm not really saying that. What I suggested for a STEP question was the additional bits and then the following statement

... With the additional parts, it's a lot more reasonable of course.

is true. I don't think anyone will solve that via the orthogonality approach unless they've already seen it.

Reply 799