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.
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}
1. What is an indeterminate form in calculus?
An indeterminate form in calculus refers to a mathematical expression that cannot be evaluated using standard algebraic techniques. These forms typically involve expressions with variables that approach certain values, such as infinity, and the result becomes undefined or ambiguous.
2. Why is the definite integral of [arcsin(1/x)-1/x] considered an indeterminate form?
The definite integral of [arcsin(1/x)-1/x] is considered an indeterminate form because the expression involves a variable, x, in the denominator that approaches 0 as x approaches infinity. This results in the expression becoming undefined, making it impossible to evaluate using traditional integration methods.
3. Is it possible to solve the definite integral of [arcsin(1/x)-1/x] without using advanced calculus techniques?
No, it is not possible to solve the definite integral of [arcsin(1/x)-1/x] using basic calculus techniques, as it is an indeterminate form. Advanced techniques such as integration by parts or substitution may be used to evaluate the integral.
4. What is the significance of the arcsine function in this integral?
The arcsine function, or inverse sine function, is significant in this integral because it is the antiderivative of the derivative of sin(x). In this case, the derivative of arcsin(1/x) results in a complex expression that cannot be easily integrated, leading to the indeterminate form.
5. Can the definite integral of [arcsin(1/x)-1/x] be approximated using numerical methods?
Yes, the definite integral of [arcsin(1/x)-1/x] can be approximated using numerical methods such as the trapezoidal rule or Simpson's rule. These methods involve dividing the integral into smaller intervals and approximating the area under the curve using geometric shapes.