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I'm interested in the integral

$$ I = \int_{1}^{\infty}\left(\frac{1}{t}-\text{arcsin}\left(\frac{1}{t}\right)\right)\text{d}t $$

I've shown that this integral exists and I know that $I = \frac{\pi}{2} - \left(1+\ln 2\right)$. I've tried multiple change of variable, for example: $$ I = \int_{0}^{1}\frac{u - \text{arcsin}\left(u\right)}{u^2}\text{d} u $$

I've then tried to $u = \sin t$ and I've obtained an integral mixing $t$ and $\sin t$ : $$ I = \lim\limits_{\epsilon \rightarrow 0}\left(\int_{\epsilon}^{1}\frac{\cos t}{\sin t}\text{d}t + \int_{\epsilon}^{1}\frac{t \cos t}{\sin^2 t}\text{d}t\right) $$ What could I try ?

MathFail
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Atmos
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1 Answers1

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$$ I = \int_{0}^{1}\frac{u - \text{arcsin}\left(u\right)}{u^2}\text{d} u $$

Integration by part: $$ I = \int_{0}^{1}\left(\arcsin(u)-u\right)d\left(\frac{1}u \right)=\arcsin(1)-1-\int_{0}^{1}\frac{\frac{1}{\sqrt{1-u^2}}-1}{u}du $$ We get: $$ I =\frac{\pi}2-1+\ln(1+\sqrt{1-u^2})\bigg|_0^1=\boxed{~\frac{\pi}2-1-\ln2~} $$

MathFail
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  • I am very sorry but I don't understand. I know how $\int_0^1\frac{u-\arcsin(u)}{u^2}du=\int_0^1(\arcsin(u)-u)d\left(\frac 1u\right)$, but I don't know what you mean by $fdg=fg-gdf$. – Kamal Saleh May 17 '23 at 20:01
  • sorry for the typo, it is $\int fdg=fg-\int g df$, and let $f=\arcsin u-u, g=1/u$ – MathFail May 17 '23 at 20:52