A Pair Of Elementary Logarithmic Integrals

Here are two beautiful elementary logarithmic integrals which I came across recently: i) Let m,n be natural numbers. Prove that01xmlogn(x)dx=(1)nn!(m+1)n+1.\textrm{i) Let} \ m, n \ \textrm{be natural numbers. Prove that} \int_{0}^1 x^m \log^n (x) \, dx=(-1)^n\frac{n!}{(m+1)^{n+1}}. ii) Prove that0axmlogn(x)dx=am+1k=0n(1)k(nk)k!(m+1)k+1×lognk(a),a>0.\textrm{ii) Prove that} \int_{0}^a x^m \log^n (x) \, dx=a^{m+1} \sum_{k=0}^n (-1)^k\binom n k \frac{k!}{(m+1)^{k+1}} \times \log^{n-k}(a) , a>0.

I'll present my own solution to i) and ii).

Solution i)

Let I(m,n)I(m,n) denote the integral, 01xmlogn(x)dx\displaystyle \int_{0}^1 x^m \log^n (x) \, dx, by using IBP, we have

I(m,n)=1m+1logn(x)xm+101mm+101xmlogn1(x)dx=nm+1I(m,n1)=(1)2n(n1)(m+1)2I(m,n2)==(1)nn!(m+1)nI(m,0)=(1)nn!(m+1)n01xmdx=(1)nn!(m+1)n+1. \begin{aligned} I(m,n)&=\left .\frac{1}{m+1} \log^n (x) x^{m+1}\right|_{0}^{1}-\frac{m}{m+1}\int_{0}^1 x^m \log^{n-1} (x) \, dx \\ &=-\frac{n}{m+1}I(m,n-1) \\ &=(-1)^2\frac{n(n-1)}{(m+1)^2} I(m,n-2) \\ &=\ldots \\ &=(-1)^{n}\frac{n!}{(m+1)^n} I(m,0) \\ &=(-1)^{n}\frac{n!}{(m+1)^n} \int_{0}^1 x^m \, dx \\ &=(-1)^n\frac{n!}{(m+1)^{n+1}}. \ \blacksquare \end{aligned}

Solution ii)

Let I(m,n)I(m,n) denote the integral, 0axmlogn(x)dx\displaystyle \int_{0}^a x^m \log^n (x) \, dx, by using IBP, we have

I(m,n)=1m+1logn(x)xm+10anm+10axmlogn1(x)dx=(1)0(n0)0!(m+1)0+1logn0(x)am+11m+1(n1)I(m,n1)=am+1(1)0(n0)0!(m+1)0+1×logn0(a)1m+1(n1)(1m+1logn1(x)am+1n1m+1I(m,n2))=am+1(1)0(n0)0!(m+1)0+1×logn0(a)+am+1(1)1(n1)1!(m+1)1+1×logn1(a)+(1)2(n2)2!(m+1)2I(m,n2)==am+1(1)0(n0)0!(m+1)0+1×logn0(a)+am+1(1)1(n1)1!(m+1)1+1×logn1(a)+am+1(1)2(n2)2!(m+1)2+1×logn2(a)++(1)n(nn)n!(m+1)nI(m,0)=am+1(1)0(n0)0!(m+1)0+1×logn0(a)+am+1(1)1(n1)1!(m+1)1+1×logn1(a)+am+1(1)2(n2)2!(m+1)2+1×logn2(a)++am+1(1)n(nn)n!(m+1)n+1×lognn(a)=am+1k=0n(1)k(nk)k!(m+1)k+1×lognk(a). \begin{aligned} I(m,n)&=\left .\frac{1}{m+1} \log^n (x) x^{m+1}\right|_{0}^{a}-\frac{n}{m+1}\int_{0}^{a} x^m \log^{n-1} (x) \, dx \\ &=(-1)^0 \binom n 0 \frac{0!}{(m+1)^{0+1}} \log^{n-0} (x) a^{m+1}-\frac{1}{m+1} \binom n 1 I(m,n-1) \\ &=a^{m+1}(-1)^0\binom n 0 \frac{0!}{(m+1)^{0+1}} \times \log^{n-0} (a)-\frac{1}{m+1} \binom n 1 \left(\frac{1}{m+1} \log^{n-1} (x) a^{m+1}-\frac{n-1}{m+1}I(m,n-2)\right) \\ &=a^{m+1}(-1)^0\binom n 0 \frac{0!}{(m+1)^{0+1}} \times \log^{n-0} (a)+a^{m+1} (-1)^1 \binom n 1 \frac{1!}{(m+1)^{1+1}} \times \log^{n-1} (a)+ (-1)^2 \binom n 2 \frac{2!}{(m+1)^2} I(m,n-2) \\ &=\ldots \\ &=a^{m+1}(-1)^0\binom n 0 \frac{0!}{(m+1)^{0+1}} \times \log^{n-0} (a)+a^{m+1} (-1)^1 \binom n 1 \frac{1!}{(m+1)^{1+1}} \times \log^{n-1} (a)+ a^{m+1} (-1)^2 \binom n 2 \frac{2!}{(m+1)^{2+1}} \times \log^{n-2} (a) +\ldots +(-1)^n \binom n n \frac{n!}{(m+1)^n}I(m,0) \\ &=a^{m+1}(-1)^0\binom n 0 \frac{0!}{(m+1)^{0+1}} \times \log^{n-0} (a)+a^{m+1} (-1)^1 \binom n 1 \frac{1!}{(m+1)^{1+1}} \times \log^{n-1} (a)+ a^{m+1} (-1)^2 \binom n 2 \frac{2!}{(m+1)^{2+1}} \times \log^{n-2} (a) +\ldots +a^{m+1}(-1)^n \binom n n \frac{n!}{(m+1)^{n+1}} \times \log^{n-n} (a) \\ &=a^{m+1} \sum_{k=0}^n (-1)^k\binom n k \frac{k!}{(m+1)^{k+1}} \times \log^{n-k}(a). \ \blacksquare \end{aligned}

Note by ChengYiin Ong
5 months, 3 weeks ago

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01xmlogn(x)dx\int_0^1 x^m \log^n(x) dx taking log(x)=t\log(x)=t and t=ut=-u 0e(m+1)ttndt=0(1)ne(m+1)uundu \int_{-∞}^0 e^{(m+1)t} t^n dt = \int_0^∞ (-1)^n e^{-(m+1)u} u^n du Taking (m+1)u=ϕ(m+1)u=\phi , then the integral becomes (1)n(m+1)n+10eϕϕndϕ=(1)nΓ(n+1)(m+1)n+1\dfrac{(-1)^n }{(m+1)^{n+1}} \int_0^∞ e^{-\phi} \phi^n d\phi = \dfrac{(-1)^n \Gamma{(n+1)} }{(m+1)^{n+1}} =n!(1)n(m+1)n+1= \dfrac{n!(-1)^n }{(m+1)^{n+1}}

Dwaipayan Shikari - 5 months, 3 weeks ago

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Here is Another way Consider I(m)=01xmdx    I(m)=1m+1I(m)= \int_0^1 x^m dx \implies I(m) = \dfrac{1}{m+1} Differentiate both sides respect to mm for nn times nI(m)mn=01xmlogn(x)dx=(1)nn!(m+1)n+1\dfrac{\partial^nI(m)}{\partial{m}^n}= \int_0^1 x^m \log^n(x) dx = \dfrac{(-1)^n n!}{(m+1)^{n+1}} A hah! :)

Dwaipayan Shikari - 5 months, 3 weeks ago

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