Any tips on how to prove this result?

$$ \int \frac{x}{x^K + c} dx = \frac{x^2 {}_2F_1 \left(1,\frac{2}{K};\frac{K+2}{K};-\frac{x^K}{c} \right)}{2c}, $$

where $${}_2F_1 (a, b;c;z)$$ is the hypergeometric function.


Assuming $K,c>0$ , we can use the geometric series to find for $|x|^K < c$ $$ \int \limits_0^x \frac{t}{t^K + c} \, \mathrm{d} t = \frac{1}{c} \sum \limits_{n=0}^\infty \left(-\frac{1}{c}\right)^n \int \limits_0^x t^{K n +1} \, \mathrm{d}t = \frac{x^2}{c} \sum \limits_{n=0}^\infty \frac{1}{Kn+2} \left(-\frac{x^K}{c}\right)^n \, .$$

Since $$ \frac{1}{K n +2} = \frac{1}{2} \frac{\frac{2}{K}}{\frac{2}{K} +n} = \frac{1}{2} \frac{\Gamma\left(\frac{2}{K}+1\right)}{\Gamma\left(\frac{2}{K}\right)} \frac{\Gamma\left(\frac{2}{K}+n\right)}{\Gamma\left(\frac{2}{K}+1+n\right)} \, ,$$ we get $$ \int \limits_0^x \frac{t}{t^K + c} \, \mathrm{d} t = \frac{x^2}{2 c} \sum \limits_{n=0}^\infty \frac{\Gamma(1+n)}{\Gamma(1)} \frac{\Gamma\left(\frac{2}{K}+n\right)}{\Gamma\left(\frac{2}{K}\right)} \frac{\Gamma\left(\frac{2}{K}+1\right)}{\Gamma\left(\frac{2}{K}+1+n\right)} \frac{\left(-x^K/c\right)^n}{n!} \, .$$ Now note that the sum on the right-hand side is exactly the definition of the hypergeometric function.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.