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I am trying to find the integral of:

$$\int_0^R \frac{bx}{x^a-b}dx$$

where $a>2$, $R$ and $b$ are constant terms.

By using Wolfram Alpha, I found out that the function does not have an antiderivative. I have also tried to solve it using integration by parts but I am not able to find the integral of the denominator term.

Can anyone please guide me to a relevant resource where similar expressions have been integrated? Also, if you have any ideas on integrating such an expression, please feel free to share them.

Thanks.

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There is an antiderivative in terms of the gaussian hypergeometric function $$\int \frac{b\,x}{x^a-b}\,dx=-\frac{1}{2} x^2 \, _2F_1\left(1,\frac{2}{a};\frac{a+2}{a};\frac{x^a}{b}\right)$$ If $a >0$ $$I_a=\int_0^R \frac{b\,x}{x^a-b}\,dx=-\frac{1}{2} R^2 \, _2F_1\left(1,\frac{2}{a};\frac{a+2}{a};\frac{R^a}{b}\right)$$ which, as usual, requires that $R^a < b$.

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    $\begingroup$ in what course if any do you learn about this function? $\endgroup$ – C Squared Aug 30 '20 at 5:48
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    $\begingroup$ @CSquared. I really don't know at which level this is explained nowdays. Do not forget that I am almost 80 ! Cheers :-) $\endgroup$ – Claude Leibovici Aug 30 '20 at 5:53
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Probably the best way of going about this one is to write it as an infinite sum, as there does not appear to be an easy closed form.

So, what we can do is write $$\int_0^R \frac{bx}{x^a-b}dx = -\int_0^R \frac{x}{1-x^a/b}dx = -\int_0^R \sum_{n=0}^\infty x\left(\frac{x^a}{b}\right)^n$$ And now the trick is that we can switch the order of integration and summation, so we get: $$-\sum_{n=0}^\infty \int_0^R x^{an+1}/b^n = -\sum_{n=0}^\infty \frac{R^{an+2}}{(n+2)b^n}$$ Note that this only works if $x^a < b$ all the time, i.e. $R^a < b$, as otherwise, we integrate across a singularity and the infinite sum diverges (another way of seeing this is $x^a=b$ causes division by $0$).

Finally, there are likely some special cases where you can get more of a closed form, but if you want a solution for all $a,b,R$ with $R^a<b$, this is probably the best we can do.

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  • $\begingroup$ Thank you so much for the help. Just to clarify the denominator of the resultant term should have $a$ in it, right? Like it should be $-\sum_{n=0}^\infty \dfrac{R^{an+2}}{(an+2)b^n}$. $\endgroup$ – shahrukh Aug 29 '20 at 18:41
  • $\begingroup$ Also if $b = jt$, where $j$ is an imaginary number and $t$ can be any value greater than 0 then would the above solution be still applicable? $\endgroup$ – shahrukh Aug 29 '20 at 19:38
  • $\begingroup$ Not if $t$ is any value greater than $0$. The issue is, if if $\dfrac{R^a}{b} = \dfrac{R^a}{jt} > 1$, we can see that this series will not converge by a sort of ratio test. $\endgroup$ – Isaac Browne Aug 31 '20 at 2:59

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