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The hard integral thread.

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Original post by DFranklin
isn't that just for sequences?

.


Yes, but many of the sequences are decimal expansions of well known (and not so well known) numbers.
Reply 381
here is a detailed solution from an integral earlier

hopefully the extra detail helps some users
This website has a rather nice problem generator for many topics:
http://i-want-to-study-engineering.org/q/frightening_integral/
Reply 383
here is a detailed solution from another integral earlier

hopefully the extra detail helps some users
Reply 384
Original post by TeeEm
here is a detailed solution from another integral earlier

hopefully the extra detail helps some users


After fiddling I got the the expansion part but that's is, really nice solution yet again TeeEm!

I haven't covered the Gamma function yet, it is similar to factorial right?



Could someone give me an integral which requires some series expansion, thanks! :biggrin:
(edited 8 years ago)
Reply 385
Original post by Louisb19
After fiddling I got the the expansion part but that's is, really nice solution yet again TeeEm!

I haven't covered the Gamma function yet, it is similar to factorial right?


gamma and beta are very nice and very easy
(they "do" it for me ...)

Learn the basics from a self teach book like advanced engineering mathematics by K Stroud
then go to my site and have practice
Reply 386
Original post by TeeEm
gamma and beta are very nice and very easy
(they "do" it for me ...)

Learn the basics from a self teach book like advanced engineering mathematics by K Stroud
then go to my site and have practice


I remember seeing it when reading about the Riemann Hypothesis.


Why don't more teachers use your resources, 10x better than those damn Solomon worksheets!
Original post by TeeEm
here is a detailed solution from another integral earlier

hopefully the extra detail helps some users


Remember this question, incredible question.



Posted from TSR Mobile
Reply 388
Original post by physicsmaths
Remember this question, incredible question.



Posted from TSR Mobile


a riddle of some sort?
Original post by TeeEm
a riddle of some sort?


Lol no, forgot the 'I' at the start, the joys of doing lots of maths.


Posted from TSR Mobile
Original post by Indeterminate
No takers? :sad: Ok, allow me to do the honours.


PRSOM. Excellent. :-)
Reply 391
01dx1x4 \displaystyle\int_0^1 \dfrac{\mathrm{d} x}{\sqrt{1 - x^4}}

Very well may be impossible.
(edited 8 years ago)
Reply 392
Original post by Louisb19
dx1x4 \displaystyle\int \dfrac{\mathrm{d} x}{\sqrt{1 - x^4}}


I have not tried it but I do not think there is a closed form without limits
Reply 393
Original post by TeeEm
I have not tried it but I do not think there is a closed form without limits


That is what I feared.
Reply 394
Original post by Louisb19
That is what I feared.
If it's from 0 to 1 then the answer is:

Spoiler

Reply 395
Original post by Kummer
If it's from 0 to 1 then the answer is:

Spoiler



That is what it was if I remember correctly. Could you post a solution?
Original post by Louisb19
dx1x4 \displaystyle\int \dfrac{\mathrm{d} x}{\sqrt{1 - x^4}}

Very well may be impossible.


This integral, or the version

0tdx1x4 \displaystyle\int_{0}^{t} \dfrac{\mathrm{d} x}{\sqrt{1 - x^4}}

lies right at the birth of the theory of elliptic functions. Gauss himself studied this integral (the inverse of which function is called the lemniscatic sine function) as a natural generalization of the sine function, which you can consider via

sin1t=0tdx1x2 \displaystyle \sin^{-1} t = \int_{0}^{t} \dfrac{\mathrm{d} x}{\sqrt{1 - x^2}}

I've linked to the Wikipedia article about it; Stillwell's excellent book "Mathematics and its History" has a fascinating account of the development of the theory.
(edited 8 years ago)
Original post by DFranklin
Google symbolic inverter
.


How did I not know this existed all of this time?!?!

Original post by Gregorius

So, in the mean-time, here's an easier one from Johann Bernoulli in the same spirit. Show that:

01xxdx=1122+133144+\displaystyle \int_{0}^{1} x^x dx = 1 - \frac{1}{2^2} + \frac{1}{3^3} - \frac{1}{4^4} + \ldots

Spoiler



This is very well known, no? Suppose it's interesting for those who haven't seen it before.

EDIT: I've been looking for an entertaining read for a long time so will give Stillwell a go, thanks for the tip :smile:
(edited 8 years ago)
Reply 398
Original post by Kummer
If it's from 0 to 1 then the answer is:

Spoiler



I might supply a neat detailed solution to that tomorrow
Reply 399
Original post by Louisb19
That is what it was if I remember correctly. Could you post a solution?
Let u=x4u = x^4 then I=1401u34(1u)12  du=1401u141(1u)121  du\displaystyle I = \frac{1}{4}\int_{0}^{1}u^{-\frac{3}{4}}(1-u)^{-\frac{1}{2}}\;{du} = \frac{1}{4}\int_{0}^{1}u^{\frac{1}{4}-1}(1-u)^{\frac{1}{2}-1} \;{du} and this is 14B(14,12)=Γ2(1/4)42π.\displaystyle \frac{1}{4} \mathrm{B}\left(\frac{1}{4}, \frac{1}{2}\right) = \displaystyle \frac{\Gamma^2(1/4)}{4\sqrt{2\pi}}.

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