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Question Number 96554 by student work last updated on 02/Jun/20

$$\int \frac{\mathrm{dx}}{\mathrm{x}!}=?$$

Commented by student work last updated on 02/Jun/20

$$\mathrm{or}\int \frac{\mathrm{x}}{\mathrm{x}!}\mathrm{dx}=?$$

Commented by MJS last updated on 03/Jun/20

$$\mathrm{I}\mathrm{stated}\mathrm{this}\mathrm{before}$$$$x!\mathrm{is}\mathrm{defined}\mathrm{for}x\in \mathbb{N}$$$$\mathrm{if}\mathrm{you}\mathrm{mean}\mathrm{\Gamma }\left(x\right)\mathrm{you}\mathrm{must}\mathrm{say}\mathrm{\Gamma }\left(x\right)$$

Answered by Rio Michael last updated on 03/Jun/20

$$\mathrm{recall}x!=\mathrm{\Gamma }(x+1)$$$$\Rightarrow \int \frac{dx}{x!}=\int \frac{1}{\mathrm{\Gamma }(x+1)}dx=$$$$x+\frac{\gamma x^2}{2}+\frac{1}{36}(6{\gamma }^2-{\pi }^2)x^3+\frac{1}{48}{x}^4(2{\gamma }^3-\gamma {\pi }^2-2{\psi }^{\left(2\right)}\left(1\right))+\frac{1}{7200}{x}^5(60{\gamma }^4+$$$$+{\pi }^4-60{\gamma }^2{\pi }^2-240\psi {\gamma }^{\left(2\right)}\left(1\right))+\frac{1}{8467420}{x}^7(168{\gamma }^4+42{\gamma }^2{\pi }^4-420{\gamma }^4{\pi }^4-3360{\gamma }^3{\psi }^{\left(2\right)}\left(1\right))$$$$-5({\pi }^6-336{\psi }^{\left(2\right)}{\left(1\right)}^{\left(2\right)})+336\gamma (5{\pi }^2{\psi }^{\left(2\right)}\left(1\right)-3{\psi }^{\left(4\right)}\left(1\right))+\frac{1}{8640}{x}^6(12{\gamma }^5+$$$$\gamma {\pi }^4-20{\gamma }^3{\pi }^2-120{\gamma }^2{\psi }^{\left(2\right)}\left(1\right)+20{\pi }^2{\psi }^{\left(2\right)}\left(1\right)-12{\psi }^{\left(4\right)}\left(1\right))+...k$$$$\gamma =\mathrm{Euler}-\mathrm{Mascheroni}\mathrm{constant}$$$${\psi }^{\left(m\right)}\left(z\right)=\mathrm{polygama}\mathrm{function}$$$$\mathrm{summarising}\mathrm{the}\mathrm{above},\mathrm{we}\mathrm{see}\mathrm{that}$$$$\int \frac{1}{\mathrm{\Gamma }(x+1)}dx=\frac{i}{2\pi }\int \left({\oint }_H{(-t)}^{-(1+x)}{e}^{-t}dt\right)dx.$$$$\mathrm{also}\mathrm{please}\mathrm{verify}\mathrm{on}\mathrm{wolframe}\mathrm{alpha}\mathrm{that}$$$$\frac{1}{\mathrm{\Gamma }(x+1)}=\underset{k=1}{\sum ^{\mathrm{\infty }}}{c}_k{(x+1)}^k,$$$$c_1=1,c_2=2\mathrm{and}{c}_k=\frac{\gamma c_{k-1}+\underset{j=2}{\sum ^{k-2}}{c}_j{(-1)}^{j+k+1}\zeta (k-j)}{k-1}$$$$\zeta \left(z\right)=\mathrm{Reiman}\mathrm{zeta}\mathrm{function}$$$$\mathrm{integrating}\mathrm{the}\mathrm{series}\mathrm{above}$$$$\int \frac{1}{\mathrm{\Gamma }(1+x)}dx=\int \left(\underset{k=1}{\sum ^{\mathrm{\infty }}}{c}_k{(x+1)}^k\right)dx=\underset{k=1}{\sum ^{\mathrm{\infty }}}{c}_k\int {(x+1)}^{k}dx$$$$\int \frac{1}{\mathrm{\Gamma }(1+x)}dx=\int \frac{dx}{x!}=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{c_k{(x+1)}^{k+1}}{k+1}+\mathrm{constant}\left(A\right)$$$$$$

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