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

lfdahl

Well-known member
Nov 26, 2013
719
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.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,714
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.
 

lfdahl

Well-known member
Nov 26, 2013
719
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!