MHB Is the definite integral ∫ [arcsin(1/x)-1/x]of indeterminate form?

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The definite integral ∫[arcsin(1/x) - 1/x] dx from 1 to infinity is analyzed for its convergence and whether it is of indeterminate form. The integrand approaches zero as x approaches infinity, suggesting potential convergence. However, a detailed evaluation is needed to determine if the integral converges or diverges. The discussion emphasizes the importance of proving the behavior of the integrand at the limits of integration. Ultimately, the conclusion regarding the integral's form hinges on rigorous mathematical proof.
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Is the definite integral

$$\int_{1}^{\infty}\left(\arcsin \left(\frac{1}{x}\right)-\frac{1}{x} \right)\,dx$$

of indeterminate form or not? Prove your statement.
 
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lfdahl said:
Is the definite integral

$$\int_{1}^{\infty}\left(\arcsin \left(\frac{1}{x}\right)-\frac{1}{x} \right)\,dx$$

of indeterminate form or not? Prove your statement.

For $x > 1$ we have the indefinite form:
\begin{aligned}\int\left(\arcsin \left(\frac{1}{x}\right)-\frac{1}{x} \right)\,dx
&= x \arcsin(1/x) - \int x\,d(\arcsin(1/x)) - \int \frac 1x\, dx \\
&= x \arcsin(1/x) - \int x\cdot \frac{1}{\sqrt{1-(1/x)^2}}\cdot -\frac 1{x^2}\,dx - \int \frac 1x\, dx \\
&= x \arcsin(1/x) + \int \frac{dx}{\sqrt{x^2-1}} - \ln x \\
&= x \arcsin(1/x) + \ln\left({\sqrt{x^2-1}} + x\right) - \ln x + C \\
&= x \arcsin(1/x) + \ln\left({\sqrt{1-1/x^2}} + 1\right) + C
\end{aligned}
Thus the improper definite integral is:
\begin{aligned}\int_1^\infty\left(\arcsin \left(\frac{1}{x}\right)-\frac{1}{x} \right)\,dx
&= \left[ x \arcsin(1/x) + \ln\left({\sqrt{1-1/x^2}} + 1\right) \right]_1^\infty\\
&= \lim_{a\to\infty} a \arcsin(1/a) + \ln 2 - \arcsin 1 \\
&= \lim_{a\to\infty}\left[ \frac{\arcsin(1/a)}{1/a} \right] + \ln 2 - \frac\pi 2 \\
&= \lim_{a\to\infty}\left[ \frac{\frac 1{\sqrt{1-1/a^2}}\cdot -\frac 1{a^2}}{-\frac1{a^2}} \right] + \ln 2 - \frac\pi 2 \\
&= 1 + \ln 2 - \frac\pi 2
\end{aligned}

Therefore the given improper definite integral is determinate.
 
I like Serena said:
For $x > 1$ we have:
\begin{aligned}\int\left(\arcsin \left(\frac{1}{x}\right)-\frac{1}{x} \right)\,dx
&= x \arcsin(1/x) - \int x\,d(\arcsin(1/x)) - \int \frac 1x\, dx \\
&= x \arcsin(1/x) - \int x\cdot \frac{1}{\sqrt{1-(1/x)^2}}\cdot -\frac 1{x^2}\,dx - \int \frac 1x\, dx \\
&= x \arcsin(1/x) + \int \frac{dx}{\sqrt{x^2-1}} - \ln x \\
&= x \arcsin(1/x) + \ln\left({\sqrt{x^2-1}} + x\right) - \ln x \\
&= x \arcsin(1/x) + \ln\left({\sqrt{1-1/x^2}} + 1\right)
\end{aligned}
Thus:
\begin{aligned}\int_1^\infty\left(\arcsin \left(\frac{1}{x}\right)-\frac{1}{x} \right)\,dx
&= \left[ x \arcsin(1/x) + \ln\left({\sqrt{1-1/x^2}} + 1\right) \right]_1^\infty\\
&= \lim_{a\to\infty} a \arcsin(1/a) + \ln 2 - \arcsin 1 \\
&= \lim_{a\to\infty}\left[ \frac{\arcsin(1/a)}{1/a} \right] + \ln 2 - \frac\pi 2 \\
&= \lim_{a\to\infty}\left[ \frac{\frac 1{\sqrt{1-1/a^2}}\cdot -\frac 1{a^2}}{-\frac1{a^2}} \right] + \ln 2 - \frac\pi 2 \\
&= 1 + \ln 2 - \frac\pi 2
\end{aligned}

Great job, I like Serena! (Nod) Thankyou for your participation!
 
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