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I've been trying to find the general form of the following integral

$$ I(x;p,q,r,a,b)=\int\frac{x^{p}\ln^{q}(x+a)}{(x+a)^{b}}dx $$ where $a,p,q \in \mathbb{N}\setminus\{0\}$ and $b \in \mathbb{Q}$. Wolfram Alpha is able to solve this for some specific values of $p$ and $q$, i.e $p=2$ and $q=2$.

I'm looking for solutions based on simple functions, like those involved in the original integral: $\ln(x+a)$, powers of $x$ or powers of the fraction $1/(x+a)^{b}$.

Any help woul be very much apreciated.

Thanks in advance.

Neves
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1 Answers1

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Using the substitution $x = t - a$, we have $$\eqalign{\int \dfrac{x^p}{(x+a)^b}\; dx &= \int \dfrac{ (t-a)^p}{t^b}\; dt = \sum_{j=0}^p {p \choose j} (-a)^{p-j} \int t^{j-b}\; dt\cr &= \sum_{j=0}^p {p \choose j} (-a)^{p-j} \dfrac{(x+a)^{j+1-b}}{j+1-b} + C} $$ (replacing $(x+a)^{j+1-b}/(j+1-b)$ by $\ln(x+a)$ if $j+1-b = 0$). Now $$\dfrac{\partial^q}{\partial b^q} (x+a)^{-b} = (-1)^q (\ln (x+a))^q(x+a)^{-b}$$ so that $$ \eqalign{\int \dfrac{x^p \ln(x+a)^q}{(x+a)^b}\; dx &= (-1)^q \sum_{j=0}^p {p \choose j} (-a)^{p-j} \dfrac{\partial^q}{\partial b^q} \dfrac{(x+a)^{j+1-b}}{j+1-b} \cr &= (-1)^q \sum_{j=0}^p \sum_{i=0}^q {p \choose j} \dfrac{q!}{i!} (-a)^{p-j} (-1)^i \dfrac{(x+a)^{j+1-b} \ln(x+a)^i}{(j+1-b)^{1+q-i}} + C}$$ This must be modified when $b$ is an integer from $1$ to $p+1$.

Robert Israel
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  • Hi, I think that there must be an error somewhere. I've tested for $\int_{0}^{1}\frac{x^2\ln^2(x+2)}{(x+2)^{1/2}}$ and your solution gives $2.40268\cdots$ but WolframAlpha gives $0.205058\cdots$ – Neves Jul 20 '15 at 13:48
  • I think the error is yours. I tried my formula, and in this case it gives $${\frac {65296,\sqrt {3}}{1125}}-{\frac {1816,\sqrt {3}\ln \left( 3 \right) }{75}}+{\frac {18,\sqrt {3} \left( \ln \left( 3 \right) \right) ^{2}}{5}}-{\frac {201728,\sqrt {2}}{3375}}+{\frac {5888, \sqrt {2}\ln \left( 2 \right) }{225}}-{\frac {64,\sqrt {2} \left( \ln \left( 2 \right) \right) ^{2}}{15}} $$ which is approximately $0.205058468$. – Robert Israel Jul 20 '15 at 15:19
  • You are right, I had a mistake. Sorry. – Neves Jul 20 '15 at 15:36
  • There is a problem when $a=1$ and $i=0$, and indeterminate of the kind $0^0$ arises. How should it be interpreted in your expression to keep everything right? – Neves Jul 20 '15 at 15:57
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    For $i=0$ we take $\ln(x+a)^0$ to be $1$.. – Robert Israel Jul 20 '15 at 19:54