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$$\int_0^c\frac{t^n}{x-t}dt$$Is there any standard formula for this?

n can be real; I need for a specific case of n=0.5. Thanks!

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    Sub $t=x+y$. Use the binomial theorem. – 2017-01-30
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    So will dt=dy then? – 2017-01-30
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    also for binomial theorem n should be an integer , what if n=1/2? – 2017-01-30

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If $n$ is a whole number, we have \begin{eqnarray} \dfrac{t^n}{x-t}&=&\dfrac{t^n-x^n}{x-t}+\dfrac{x^n}{x-t}=t^{n-1}+xt^{n-2}+\ldots+x^{n-2}t+x^{n-1}-x^n\cdot\dfrac{1}{t-x}\\ &=&\sum_{k=0}^{n-1}x^{n-1-k}t^k-x^n\cdot\dfrac{1}{t-x}, \end{eqnarray} it follows that \begin{eqnarray} \int_0^c\dfrac{t^n}{x-t}\,dt&=&\sum_{k=0}^{n-1}x^{n-1-k}\int_0^c t^k\,dt-x^n\cdot\int_0^c\dfrac{1}{t-x}\,dt=\sum_{k=0}^{n-1}x^{n-1-k}\dfrac{t^{k+1}}{k+1}\Big|_{t=0}^{t=c}-x^n\cdot\ln|t-x|\Big|_{t=0}^{t=c}\\ &=&\sum_{k=0}^{n-1}\dfrac{c^{k+1}}{k+1}x^{n-1-k}-x^n[\ln|x-c|-\ln|x|] \end{eqnarray}

If $n=0.5$, set $t=u^2$, so that $$ I(c,x)\int_0^c\dfrac{t^{0.5}}{x-t}\,dt=\int_0^{\sqrt{c}}\dfrac{u}{x-u^2}\,du $$ Clearly $$ I(c,0)=-\int_0^{\sqrt{c}}\dfrac{1}{u}\,du=-\ln\sqrt{c}=-\dfrac12\ln c. $$ If $x\ne0$, then $$ I(c,x)=-\int_0^{\sqrt{c}}\dfrac{u}{u^2-x}\,du=-\dfrac12\ln|u^2-x|\Big|_{u=0}^{u=\sqrt{c}}=-\frac12\left(\ln|x-c|-\ln|x|\right) $$

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    Is this true for a fractional n? – 2017-01-30
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    If we use your method for sa n=1, the first term in the summation is +c, where as the actual answer is -c. – 2017-01-31
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    ahh, I got it, when you use a^n-b^n expansion, you will have (a-b) in the numerator, but we have (b-a) in the denominator, so there will be negative sign in front of the series expansion. You can edit that. Thanks a lot! – 2017-01-31