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Solve the following integral $$\int_0^1 x^4\left(\ln\frac{1}{x}\right)^3dx$$

my attempt:

$$\int_0^1 x^4\left(\ln\frac{1}{x}\right)^3dx$$ $$=-\int_0^1 x^4\left(\ln x\right)^3dx$$ substitute $ln x=t, \ dx=e^tdt$

$$=-\int_{-\infty}^0 (e^t)^4t^3e^t\ dt$$ $$=-\int_{-\infty}^0 t^3e^{5t}\ dt$$ $$=\int_0^{-\infty} t^3e^{5t}\ dt$$ I got stuck here. I am not sure how to proceed. please help me solve it.

Thanks.

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  • $\begingroup$ I think we need to use the upper incomplete gamma function. Just substitute $5t=-u$, and everything should work out $\endgroup$ Oct 24, 2021 at 4:55
  • $\begingroup$ Integrate by parts 3 times. $\endgroup$
    – amsmath
    Oct 24, 2021 at 4:56

2 Answers 2

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by using ${\int}\mathtt{f}\mathtt{g}' = \mathtt{f}\mathtt{g} - {\int}\mathtt{f}'\mathtt{g} ,f=(ln)^3,g=x^4$ $$=\int_0^1 x^4\left(\ln x\right)^3dx=\dfrac{x^5\ln^3\left(x\right)}{5}-{\displaystyle\int}\dfrac{3x^4\ln^2\left(x\right)}{5}\,\mathrm{d}x $$ using integration by parts $${\displaystyle\int}\dfrac{3x^4\ln^2\left(x\right)}{5}\,\mathrm{d}x =\frac 35 (=\dfrac{x^5\ln^2\left(x\right)}{5}-{\displaystyle\int}\dfrac{2x^4\ln\left(x\right)}{5}\,\mathrm{d}x )$$ and once again

$${\displaystyle\int}x^4\ln^3\left(x\right)\,\mathrm{d}x =\\ =\dfrac{x^5\ln^3\left(x\right)}{5}-\dfrac{3x^5\ln^2\left(x\right)}{25}+\dfrac{6x^5\ln\left(x\right)}{125}-\dfrac{6x^5}{625} $$

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You are almost done.

Substitute $5t=-y\implies dt=-\frac{dy}{5}$ $$\begin{align*} \int_0^{-\infty} t^3e^{5t}\ dt &=\int_0^{\infty} \left(-\frac y5\right)^3e^{-y}\ \left(-\frac{dy}{5}\right)\\ &=\frac{1}{625}\int_0^{\infty} e^{-y}y^3dy\quad (\text{Recall Gamma \\Function})\\ &=\frac{1}{625}\Gamma{4}\\ &=\frac{6}{625}\\ \end{align*}$$

Substitute $5t=-y\implies dt=-\frac{dy}{5}$ $$\begin{align*} \int_0^{-\infty} t^3e^{5t}\ dt &$=\int_0^{\infty} \left(-\frac y5\right)^3e^{-y}\ \left(-\frac{dy}{5}\right)\\ &=\frac{1}{625}\int_0^{\infty} e^{-y}y^3dy\quad (\text{Recall Laplace Transform})\\ &=\frac{1}{625}L\left[t^3\right]_{s=1}\quad \left(\because L[t^n]=\int_0^{\infty}e^{-st}t^n\ dt=\frac{\Gamma(n+1)}{s^{n+1}}\right)\\ &=\frac{1}{625}\left[\frac{\Gamma{4}}{s^4}\right]_{s=1}\\ &=\frac{\Gamma{4}}{625}\\ &=\frac{6}{625}\\ \end{align*}$$

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