Sejam \(m,n\) naturais, e a um número real, então vale a igualdade: \[\int_{0}^{a}x^m\log ^n(x)dx=a^{m+1}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{k!\log^{n-k}a}{(m+1)^{k+1}}\]
Proof. Defina:
\[I_{n,m}=a^{m+1}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{k!\log^{n-k}a}{(m+1)^{k+1}}\]
Daí então, observe que:
\[I_{n,m}-\frac{a^{m+1}\log^na}{m+1}=a^{m+1}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{k!\log^{n-k}a}{(m+1)^{k+1}}-\frac{a^{m+1}\log^na}{m+1}= a^{m+1}\sum_{k=1}^{n}(-1)^k{n \choose k}\frac{k!\log^{n-k}a}{(m+1)^{k+1}}=\]
\[a^{m+1}\sum_{k=0}^{n-1}(-1)^{k+1}{n \choose k+1}\frac{(k+1)!\log^{n-(k+1)}a}{(m+1)^{k+2}}= \frac{a^{m+1}}{m+1}\sum_{k=0}^{n-1}(-1)^{k+1}{n \choose k+1}\frac{(k+1)!\log^{n-(k+1)}a}{(m+1)^{k+1}} \ (*)\] Veja que \(\displaystyle {n \choose k+1}(k+1)!=\frac{n!(k+1)!}{(k+1)!(n-(k+1))!}=n\frac{(n-1)!k!}{k!(n-1-k)!}=n{n-1 \choose k}k!\), substituindo logo acima, teremos:
\[(*) \ \frac{a^{m+1}}{m+1}\sum_{k=0}^{n-1}(-1)^{k+1}{n \choose k+1}\frac{k+1!\log^{n-(k+1)}a}{(m+1)^{k+1}}= \frac{a^{m+1}}{m+1}\sum_{k=0}^{n-1}(-1)^{k+1}n{n-1 \choose k}\frac{k!\log^{n-(k+1)}a}{(m+1)^{k+1}}=\]
\[-\frac{a^{m+1}n}{m+1}\sum_{k=0}^{n-1}(-1)^{k}{n-1 \choose k}\frac{k!\log^{n-1-k}a}{(m+1)^{k+1}}=-\frac{n}{m+1}I_{n-1,m}\Rightarrow I_{n,m}-\frac{a^{m+1}\log^na}{m+1}=-\frac{n}{m+1}I_{n-1,m} \Rightarrow\]
\[I_{n,m}=\frac{a^{m+1}\log^na}{m+1}-\frac{n}{m+1}I_{n-1,m}\] Seja \(\displaystyle I'_{n,m}=\int_{0}^{a} x^m\log^n xdx\) escolha \(\displaystyle u=\log^nx,dv=x^mdx\) e veja que, por integração por partes teremos:
\[I'_{n,m}=\frac{a^{m+1}\log^na}{m+1}-\lim_{x\rightarrow 0}\frac{x^{m+1}\log^nx}{m+1}-\frac{n}{m+1}I'_{n-1,m}=\frac{a^{m+1}\log^na}{m+1}-\frac{n}{m+1}I'_{n-1,m}\Rightarrow\] \[I'_{n,m}=\frac{a^{m+1}\log^na}{m+1}-\frac{n}{m+1}I'_{n-1,m}\] As sequências tem a mesma relação recursiva, e desde que se tome como valor inicial \(n=0\) observa-se que \(\displaystyle I_{n,m}=I'_{n,m}\). ◻
Defina os operadores logo abaixo: \[\nabla_{m,k}:= \int_{0}^{x_{1,k}}\int_{0}^{x_{2,k}}\int_{0}^{x_{3,k}}....\int_{0}^{x_{m-1,k}}\int_{0}^{x_{m,k}}\]
\[\delta_{m,k}:=dx_{1,k} \ dx_{2,k} \ dx_{3,k} \ dx_{4,k}....dx_{m,k}\]
Defina as funções fundamentais \(\displaystyle H_1:\mathbb{R}^n\rightarrow \mathbb{R}\) e \(\displaystyle G_1:\mathbb{R}^n\rightarrow \mathbb{R}\) com suas respectivas leis de formação: \[H_1(x_{u,v}):=\frac{1}{x_{u,v}^u}\nabla_{u,v}....\frac{1}{x_{l,2}^l}\nabla_{l,2}\frac{1}{x_{j,1}^j}\nabla_{j,1}\frac{w^n}{(e^{1/w})^{m+1}} \int_{0}^{e^{1/w}}x^m\log ^{n}(x) \ dx \ dw \ \delta_{j,1} \ \delta_{l,2} \ \delta_{o,4} \ \delta_{p,5} \ \delta_{q,6}...\delta_{u,v}\]
\[G_1(w):=\frac{(m+1)w^n}{(e^{1/w})^{m+1}} \int_{0}^{e^{1/w}}x^m\log ^{n}(x)dx\]
Defina as sequências de funções: \[H_k(x_{u,v}):=j_kH_{k-1}(x_{u,v})- x_{u,v}H'_{k-1}(x_{u,v}) \ \ ; \ \ G_k(w):=j_kG_{k-1}(w)- wG'_{k-1}(w)\] Onde \(\displaystyle j_k, k \in \mathbb{N}\).Então existe uma função \(\displaystyle I_k(x_{u,v})\) e uma constante \(\vartheta_k\) tal que:
\[H_k(x_{u,v})+\vartheta_kI_{k-1}(x_{u,v})=G_k(w)+x_{u,v}I'_{k-1}(x_{u,v})\]
Defina um objeto como sendo um elemento qualquer de uma das sequências contados do valor inicial da sequência até que ocorra \(\displaystyle H=G=0\).Mostrar que existem exatamente \(\displaystyle n!\) vezes de sequenciar objetos distintos de \(\displaystyle G\) e \(\displaystyle H\).Prove que:
\(\displaystyle {(-1)}^n\frac{n!H_n(x)(j_1-n)(j_2-n)...(j_\varphi-n)}{{(m+1)}^{m+1}}+\sum_{k=1}^{n-1}{H_k(x){(-1)}^k}\binom{n}{k}\frac{k!(j_1-k)(j_2-k)...(j_\varphi-k)}{{(m+1)}^{k+1}}\)
\[-\sum_{k=1}^{n-1}{(-1)}^{k+1}{H_k(x)\binom{n}{k+1}\frac{(k+1)!x^{k+1}(j_1-(k+1))(j_2-(k+1))(j_3-(k+1))...(j_{{\varphi}}-(k+1))}{{(m+1)}^{k+2}}}=\]
