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 that∫01xmlog⁡n(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 that∫0axmlog⁡n(x) dx=am+1∑k=0n(−1)k(nk)k!(m+1)k+1×log⁡n−k(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, ∫01xmlog⁡n(x) dx\displaystyle \int_{0}^1 x^m \log^n (x) \, dx, by using IBP, we have

I(m,n)=1m+1log⁡n(x)xm+1∣01−mm+1∫01xmlog⁡n−1(x) dx=−nm+1I(m,n−1)=(−1)2n(n−1)(m+1)2I(m,n−2)=…=(−1)nn!(m+1)nI(m,0)=(−1)nn!(m+1)n∫01xm dx=(−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, ∫0axmlog⁡n(x) dx\displaystyle \int_{0}^a x^m \log^n (x) \, dx, by using IBP, we have

I(m,n)=1m+1log⁡n(x)xm+1∣0a−nm+1∫0axmlog⁡n−1(x) dx=(−1)0(n0)0!(m+1)0+1log⁡n−0(x)am+1−1m+1(n1)I(m,n−1)=am+1(−1)0(n0)0!(m+1)0+1×log⁡n−0(a)−1m+1(n1)(1m+1log⁡n−1(x)am+1−n−1m+1I(m,n−2))=am+1(−1)0(n0)0!(m+1)0+1×log⁡n−0(a)+am+1(−1)1(n1)1!(m+1)1+1×log⁡n−1(a)+(−1)2(n2)2!(m+1)2I(m,n−2)=…=am+1(−1)0(n0)0!(m+1)0+1×log⁡n−0(a)+am+1(−1)1(n1)1!(m+1)1+1×log⁡n−1(a)+am+1(−1)2(n2)2!(m+1)2+1×log⁡n−2(a)+…+(−1)n(nn)n!(m+1)nI(m,0)=am+1(−1)0(n0)0!(m+1)0+1×log⁡n−0(a)+am+1(−1)1(n1)1!(m+1)1+1×log⁡n−1(a)+am+1(−1)2(n2)2!(m+1)2+1×log⁡n−2(a)+…+am+1(−1)n(nn)n!(m+1)n+1×log⁡n−n(a)=am+1∑k=0n(−1)k(nk)k!(m+1)k+1×log⁡n−k(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}

#Calculus

Note by ChengYiin Ong
4 months, 1 week ago

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Comments

∫01xmlog⁡n(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+1∫0∞e−ϕϕndϕ=(−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 - 4 months, 1 week ago

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=∫01xmlog⁡n(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 - 4 months, 1 week ago
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