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$$\int_{0}^{\infty}e^{-1\over x}\sin\left({1\over x}\right)\ln(x){\mathrm dx\over x^{4n+1}}=(-1)^{n+1}\pi F(n)\tag1$$

$n\ge1$

$F(1)={3\over 8}$

$F(2)={315\over 4}$

$F(3)=155925$

I can't seem to figure the pattern for $F(n)$

How can we evaluate the closed form for $(1)?$


$$\ln(x)=\sum_{n=1}^{\infty}{1\over n}\left({x-1\over x}\right)^n$$

$$\sum_{n=1}^{\infty}{1\over m}\int_{0}^{\infty}e^{-1\over x}\sin\left({1\over x}\right)(x-1)^m{\mathrm dx\over x^{4n+m+1}}\tag2$$

$u={1\over x}$

$$\sum_{n=1}^{\infty}{1\over m}\int_{0}^{\infty}e^{-u}\sin(u)(u^{-1}-1)^m{\mathrm du\over (u^{-1})^{4n+m-1}}\tag3$$

$$\sum_{m=1}^{\infty}{1\over m}\int_{0}^{\infty}e^{-u}\sin(u)(1-u)^m{\mathrm du\over u^{1-4n}}\tag4$$

$$\sum_{m=1}^{\infty}{1\over m}\sum_{k=0}^{m}(-1)^k{m\choose k}\int_{0}^{\infty}e^{-u}\sin(u){\mathrm du\over u^{1-4n-k}}\tag5$$


$$\boxed{\int_{0}^{+\infty} x^{4n-1} e^{-x^n}\sin(x^n)\log(x)\,dx=\color{blue}{-\frac{3\pi}{8n^{2}}}.} $$

sligth variation to

$$\boxed{\int_{0}^{+\infty} x^{4n-1} e^{-x}\sin(x)\log(x)\,dx=\color{blue}{\pi(-1)^n\frac{(4n-1)!}{4^{n+1}}}.} $$

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  • $\begingroup$ Feynman's trick and the $\Gamma$ function solve this problem in a pretty neat way. $\endgroup$ – Jack D'Aurizio Oct 26 '17 at 20:43
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$$\int_{0}^{+\infty} x^{4n-1} e^{-x}\sin(x)\log(x)\,dx=\left.\frac{d}{dm}\int_{0}^{+\infty}x^m e^{-x}\sin(x)\,dx\right|_{m=4n-1} $$ and $$ \int_{0}^{+\infty}x^m e^{-x}\sin(x)\,dx = \text{Im}\int_{0}^{+\infty} x^m e^{-(1-i)x}\,dx=\frac{\Gamma(m+1)}{\sqrt{2}^{m+1}}\sin\left(\frac{\pi}{4}(m+1)\right) $$ hence $$\boxed{\int_{0}^{+\infty} x^{4n-1} e^{-x}\sin(x)\log(x)\,dx=\color{blue}{\pi(-1)^n\frac{(4n-1)!}{4^{n+1}}}.} $$

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