My friend asked me how to integrate the following: $$\int_1^e \frac{dx}{x(x+(\ln x)^2)}$$
How am I going to solve this?Any help is greatly appreciated. Thanks.
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2This strongly suggests the substitution $u = \ln x$. – Jul 14 '14 at 05:09
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5@T.Bongers The resulting integral is still fairly challenging IMO, and makes me wonder if the problem should actually read $\int_1^e \frac{dx}{x(1+(\ln x)^2)}$. – David H Jul 14 '14 at 05:13
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1@DavidH Perhaps; it doesn't look like there's a closed form for this integral. – Jul 14 '14 at 05:21
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Maple doesn't give an answer, except when asked to do it numerically; then it gives 0.5657522781. (When I ask for 50 digits, I get 0.56575227809794093284778647981402016035012640346360). – Gerry Myerson Jul 14 '14 at 07:08
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Have you checked, Philip, whether @David is right? – Gerry Myerson Jul 16 '14 at 04:00
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Assume $u=\ln x$, then \begin{align} I:=\int_1^e \frac{dx}{x(x+(\ln x)^2)} =\int_0^1 \frac{e^u du}{e^u(e^u+u^2)} &=\int_0^1 \frac{ du}{e^u+u^2} \\ &=\int_0^1 \frac{e^{-u} du}{1+u^2e^{-u}} \end{align} Using geometric series $$\frac{1}{1+u^2e^{-u}}=\sum_{n=0}^{\infty}{(-1)^nu^{2n} e^{-nu} }$$ term by term integration, we have \begin{align} I=\sum(-1)^n (\int_0^1 { u^{2n} e^{-(n+1)u}du}=J_n) \end{align} Integrating by parts we get (from tables) $$J_n=-e^{-(n+1) u}\sum_{k=0}^{2n}\frac{(2n)!}{(2n-k)!}(n+1)^{-(k+1)}u^{2n-k}$$
Mohammad W. Alomari
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1If $J_n$ stands for an integral, then it should not depend on $u$. Is your $J_n$ a function of $u$ and you mean $J_n(1) - J_n(0)$ in your expression of $I$? – achille hui Jul 26 '14 at 06:43
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Yes exactly, l post from a smart phone so it is difficult on me to print. – Mohammad W. Alomari Jul 26 '14 at 07:43