Solving such integrals numerically?

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Discussion Overview

The discussion centers around the integral \(\int_{0}^{1} \frac{1}{\sqrt{100-x^5}} dx\), exploring whether it can be solved analytically and how to approach it numerically if not. Participants share various methods and perspectives on numerical integration techniques.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the possibility of solving the integral analytically and seeks numerical methods as alternatives.
  • Another participant suggests using finite Riemann sums, mentioning the Trapezoid rule and Simpson's rule as potential numerical methods.
  • A participant claims that the integral can be expressed in terms of a hypergeometric function, providing a specific form according to Maple.
  • Another participant presents a similar expression from Mathematica, noting a discrepancy in the parameters of the hypergeometric function and questioning its significance.
  • One participant expresses frustration over the lack of analytical solutions for integrals, suggesting a desire for a unified theory for solving such problems.
  • Another participant appreciates the results expressed in terms of special functions, indicating a preference for non-elementary mathematics.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the analytical solvability of the integral, with some expressing skepticism about finding a closed form, while others provide different representations of the integral using hypergeometric functions. The discussion remains unresolved regarding the significance of the differences in the hypergeometric parameters.

Contextual Notes

There are unresolved questions regarding the equivalence of the hypergeometric function representations provided by different software, as well as the limitations of numerical methods in achieving high precision.

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Is there any way of solving [tex]\int_{0}^{1} \frac{1}{\sqrt{100-x^5}} dx[/tex] by some regular method? If not, how does one go about solving such integrals numerically?
 
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1. I don't know about any method to crack this integral analytically.
2. As for numerical methods:
These typically start with some fancy finite Riemann sum to approximate the integral.
You could check up on the Trapezoid rule or Simpson's rule or some other rule..
 
[tex]\int_{0}^{1} \frac{dx}{\sqrt{100-x^{5}}} =\frac{1}{10} \ _{2}F_{1}\left(\frac{1}{2},\frac{1}{5},\frac{6}{5},\frac{1}{100}\right)[/tex]

according to my friend,Maple.

Daniel.
 
According to Mathematica:

[tex]\int_{0}^{1} \frac{dx}{\sqrt{100-x^{5}}} = \frac{1}{10} \text{Hypergeometric2F1} \left[ \frac{1}{5},\frac{1}{2},\frac{6}{5},\frac{1}{100} \right][/tex]

Hmm wonder why the fractions are different, does it make a difference?

Anyway to 100 s.f:

0.1000836762086654607634453194543998535914445494139945308954127536431450557575506273354667189780577481
 
Last edited:
Conventions,on normal basis they should be the same...

Gauss's function is [itex]_{2}F_{1}\left(a,b,c,z\right)[/itex]

Daniel.


P.S.Here are 125 sig.digs.

[tex]\allowbreak .\,10008\,36762\,08665\,46076\,34453\,19454\,39985\,35914\,44549\,41399\,45308\,95412\,[/tex]
[tex]75364\,31450\,55757\,55062\,73354\,66718\,97805\,77480\,65696\,66047\,63945\,64052\,7984[/tex]
 
Last edited:
dextercioby said:
P.S.Here are 125 sig.digs.

[tex]\allowbreak .\,10008\,36762\,08665\,46076\,34453\,19454\,39985\,35914\,44549\,41399\,45308\,95412\,[/tex]
[tex]75364\,31450\,55757\,55062\,73354\,66718\,97805\,77480\,65696\,66047\,63945\,64052\,7984[/tex]
Ahh so it's going to be like that is it, well in that case, to 2000 sf:

0.1000836762086654607634453194543998535914445494139945308954127536431450557575\
506273354667189780577480656966604763945640527983993080671953023089297911515248\
996013282637274784238797044064349870623646876674724483462693995830072965962405\
190019049666775160232759330205668592578946075338958508523904278007888848833088\
060488556739935166104773509141768174230265551654691550416849059337448757791416\
602630567642386763728600302390370505747803022415391215722818064179823297750442\
434640467815965016423148168846638958221953928239081869470007493153876431418021\
157217678724106639004941740270189408736496188721513628469894013953107138449253\
791957534610058071313307798195976366400486448615041008866156544661543944553681\
807447545511569915551726649431123256963503190869480487316815727122678140209122\
738110692425545202808759680890972292418663402273431929145662980727173557540648\
152030821410027920261779677495993071187486068413623093242779260125263946810946\
539669010663042017697120941744353678783323221747296966263595416822115595315497\
544773344894826763667430935304194135002414853743041716586627006645641572144003\
038754961014790401073888263046103372397829209944018459246323194799313562020937\
625258141676770207971455031095503662945181184143286220172118052176868976940819\
363006570343110478540840140496393128738996412270357183445395020573388989278555\
813556144036563102455032850874625946333692746942852322508699926140544281120662\
532525325402041764035161169228790566576239939498698256797892476673581793593341\
565591033055741801465791277510653241163913554713152060127185501516632153831744\
084485276203982058024033779697637618294711132918471464969716668181939944429129\
590221571580066039538720668949128150132064065578982106579514989803518960288881\
581003421562857862205496312407174042287815539537348909650284394685560662915042\
685107335399595953096758002153118745564316408031460221673554730734019949375781\
303013799317984319741189620338837708872894884720821818310366605811188532339923\
2084126619627384173536035029500741646467671020765461

:-p :-p :-p
 
Heh,i use a 10 y.o.Maple.It can't more than 1000 decimals.:frown:

Daniel.
 
It always breaks my heart if an integral has no analytical solution. I hate it when those damned Maple or Mathematica-programs are needed in order to crack the problem. Doesn't it disturbe you that we don't have a TOE when it come to integrals. You know, how about a String Theory-variant for integral-solving...i dunno, just wondering...

regards
marlon
 
I'm not pissed off.I like when the results come out in terms of "fancy functions" (i call them:"common special functions").

This is not really elementary mathematics...

Daniel.
 

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