How do you analytically evaluate $$\int_{0}^{\infty}\frac{x^r(-\log(1-\frac{x}{c}))^n}{c-x}dx$$ where $c,r$ are real constants, and $n$ is an integer. I've tried running it on Wolfram-alpha but the time limit exceeds. Is there a way to analytically approach it? How do I even argue that it exists?!
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1$\begingroup$ First : your left and right parentheses do not match. Second : are you sure about the bounds ? May be $c <0$; otherwise, I see many problems here. $\endgroup$– Claude LeiboviciOct 8, 2017 at 6:20
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$\begingroup$ @ClaudeLeibovici Fixed the paranthesis, and I actually copied the integral wrong! So I just corrected it. Sorry for the trouble $\endgroup$– AspiringMatOct 8, 2017 at 6:47
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$\begingroup$ Where does this problem come from ? $\endgroup$– Claude LeiboviciOct 8, 2017 at 14:00
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$\begingroup$ I stumbled across it on Facebook... $\endgroup$– AspiringMatOct 9, 2017 at 6:18
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2$\begingroup$ A substitution $y=-\log (1-x/c)$ may be useful. $\endgroup$– Alex RavskyOct 17, 2017 at 8:42
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2 Answers
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First note the following:
- This integral does not exist if $r\geq 0$, since the decay at infinity is insufficient.
- Similarly, $r>-1$ because otherwise the singularity in 0 is too large.
- Substituting $y=x/c$, you may assume that $c=1$ or $c=-1$. For $c=1$ the integral does not exist since $\frac{1}{x-1}\log(x-1)^n = \frac{d}{dx} \frac{\log(x-1)^{n+1}}{n+1}$ diverges as x tends to 1.
So instead of the integral you are probably searching for an anti-derivative. By the above considerations it suffices to consider (with $\pm 1$)
$\int x^r \frac{\log(x+1)^n}{x+1} dx$
Integrating the second factor by parts this leads to
$\int x^{r-1} \log(x+1)^n dx$
If you now additionally assume that $r$ is an integer you can substitute $x+1$ and reduce to considering
$\int x^m \log(x)^n dx$
Here, you substitute again by $x=e^y$ and obtain:
$\log ^{n+1}(x) (-(m+1) \log (x))^{-n-1} (-\Gamma (n+1,-(m+1) \log (x)))$
TLDR: You want more assumptions and should use integration by parts and substitution.
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$\begingroup$ Hint, e.g: $\,\displaystyle\int\limits_0^\infty \frac{(-\ln(1-(-x)))^2}{\sqrt{x}(-1-x)}dx\approx -16.373\enspace$ -> $\enspace$ If you meant with the second point, that it has to be $\,r\leq -1\,$ then your argumentation is wrong. $\endgroup$ Oct 23, 2017 at 9:57
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$\begingroup$ In your example r=-1/2, which is between -1 and 0, as claimed. So I am not understanding your hint. $\endgroup$ Dec 8, 2017 at 22:48
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$\begingroup$ With the first point you say that the integral doesn't exist. With the second point it seems to me that perhaps you mean that for $-1<x<0$ the integral doesn't exist too - but I was not sure. Therefore I wrote "IF you meant ..." . The second point should be made clearer, that's all. $\endgroup$ Dec 9, 2017 at 13:12
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Here is some notes. With substitution $x=c\dfrac{t}{t-1}$ we obtain $$I=\int_{0}^{\infty}\frac{x^r(-\ln(1-\frac{x}{c}))^n}{c-x}dx=\dfrac{c^r}{(-1)^{r+1}}\int_0^1t^r(1-t)^{-r-1}\ln^n(1-t)\,dt$$ and with substitution $1-t=e^{-u}$ \begin{align} I &= c^r(-1)^{n+r+1}\int_0^\infty(e^{u}-1)^ru^n\,du \\ &= c^r(-1)^{n+r+1}\int_0^\infty(1-e^{-u})^re^{ur}u^n\,du \\ &= c^r(-1)^{n+r+1}\int_0^\infty e^{ur}u^n\,du\sum_{k=0}^\infty\dfrac{\Gamma(-r+k)}{\Gamma(-r)}\dfrac{(e^{-u})^k}{k!} \\ &= c^r(-1)^{n+r+1}\sum_{k=0}^\infty\dfrac{\Gamma(-r+k)}{\Gamma(-r)k!}\int_0^\infty e^{-u(k-r)}u^n\,du \end{align} for $r<0$ (only) we see \begin{align} I &= c^r(-1)^{n+r+1}\sum_{k=0}^\infty\dfrac{\Gamma(k-r)}{\Gamma(-r)k!(k-r)^{n-1}}\Gamma(n+1) \\ &= c^r(-1)^{n+r+1}\dfrac{\Gamma(n+1)}{\Gamma(-r)}\sum_{k=0}^\infty\dfrac{\Gamma(k-r)}{k!(k-r)^{n-1}} \end{align} which is a closed form of integral.
