I am looking for idea to solve below integral

$$ \int_{0}^{a}\frac{{4\pi}\cos \left(\frac{\pi x}{30}\right)}{3\sqrt[3]{1-\left(\frac xa\right)^{3/4}} } \mathrm dx$$

I tried to use substitution, but got stuck

Can someone help me to find this integral. I am thankful for you hint in advanced. (It was in thermodynamics book)

  • 1
    $\begingroup$ at least this beomes uglier then a bessel function $\endgroup$ – tired Jun 21 '17 at 19:43

Of course, you first want to do a change of variable $u = x/a$. For positive integers $j$,

$$ \int_0^1 \frac{x^j\; dx}{\left(1-x^{3/4}\right)^{1/3}} = {\frac {4\;\Gamma \left( 4(j+1)/3 \right) \Gamma \left( 2/3 \right) }{3\;\Gamma \left( (4j+6)/3\right) }} $$

Then $ \int_0^1 \frac{\cos(tx)\; dx}{(1-x^{3/4})^{1/3}}$ can be expressed as a rather nasty combination of hypergeometric functions: $$ \eqalign{\int_0^1 \frac{\cos(tx)\; dx}{(1-x^{3/4})^{1/3}} &= {\frac {8\,\pi\,\sqrt {3}}{27} {\mbox{$_6$F$_{11}$}\left({\frac{7}{24}},{\frac{5}{12}},{\frac{13}{24}},{\frac{19}{24}},{\frac{11}{12}},{\frac{25}{24}};\frac14,\frac13,\frac38,\frac12,\frac12,\frac58,\frac34,\frac56,{\frac{7}{8}},1,{\frac{9}{8}};\,{\frac {-{t}^{6}}{46656}}\right)} }\cr &- {\frac {81\,{t}^{2}}{220} {\mbox{$_7$F$_{12}$}\left(\frac58,\frac34,{\frac{7}{8}},1,{\frac{9}{8}},\frac54,{\frac{11}{8}};\,{\frac{7}{12}},\frac23,{\frac{17}{24}},\frac56,\frac56,{\frac{23}{24}},{\frac{13}{12}},\frac76,{\frac{29}{24}},\frac43,\frac43,{\frac{35}{24}};\,{\frac {-{t}^{6}}{46656}}\right)} }\cr &+{\frac {187\,\sqrt {3} \Gamma \left( \frac23 \right)^{3}{ t}^{4}}{11856\,\pi} {\mbox{$_6$F$_{11}$}\left({\frac{23}{24}},{\frac{13}{12}},{\frac{29}{24}},{\frac{35}{24}},{\frac{19}{12}},{\frac{41}{24}};\,{\frac{11}{12}},{\frac{25}{24}},\frac76,\frac76,{\frac{31}{24}},{\frac{17}{12}},\frac32,{\frac{37}{24}},\frac53,\frac53,{\frac{43}{24}};\,{\frac {-{t}^{6}}{46656}}\right)} } }$$ I don't know if this can be simplified.

  • $\begingroup$ 1:thank you . 2:the answer does not depend on $a$ ? $\endgroup$ – Khosrotash Jun 22 '17 at 4:02
  • $\begingroup$ It does. For one thing, $t$ will depend on $a$. Also there's a factor of $a$ from the change of variables. $\endgroup$ – Robert Israel Jun 22 '17 at 4:25

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