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Find a value of C so that the given improper integral converges

Find a value for the constant C \in \mathbb{R} so that the improper integral

    \[ \int_0^{\infty} \left( \frac{1}{\sqrt{1+x^2}} - \frac{C}{x+1} \right) \, dx \]

converges and find the value of the integral in this case.


First, we have

    \begin{align*}  \int_0^{\infty} \left( \frac{1}{\sqrt{1+2x^2}} - \frac{C}{x+1} \right) \, dx &= \int_0^{\infty} \frac{x+1-C\sqrt{1+2x^2}}{(x+1)\sqrt{1+2x^2}} \, dx \\[9pt]  &= \int_0^{\infty} \frac{x+1-xC\sqrt{2 + \frac{1}{x^2}}}{(x^2+x)\sqrt{2 + \frac{1}{x^2}}} \, dx \\[9pt]  &= \int_0^{\infty} \frac{x \left(1 - C \sqrt{2 + \frac{1}{x^2}} \right) + 1}{x^2 \sqrt{2 + \frac{1}{x^2}} + x \sqrt{2 + \frac{1}{x^2}}} \, dx. \end{align*}

For this to converge we must have the coefficient of the x in the numerator going to 0 as x \to \infty (otherwise the integral will converge by limit comparison with \int \frac{1}{x}). Hence,

    \[ \lim_{x \to \infty} \left(1 - C \sqrt{2 + \frac{1}{x^2}} \right) = 1 - \sqrt{2} C \quad \implies \quad C = \frac{\sqrt{2}}{2}. \]

Now, we need to evaluate the integral. Incomplete.

One comment

  1. Tyler says:

    No idea how to use LaTeX on this blog, so I’ll just try to be as unambiguous as I can.

    The expression on the left has the indefinite integral (arcsinh(sqrt(2)x))/(sqrt(2)), and the ind. integral of the expression on the right is just log(x+1)/sqrt(2), and we simply subtract the second from the first to get the whole integral:

    (from 0 to a) integral of (1/sqrt2)*(arcsinh(sqrt(2)x)- log(x+1)) is just
    (1/sqrt2)*(arcsinh(sqrt(2)a)- log(a+1)) [since arcsinh(0) = 0, and obv. ln(1) = 0]
    Given the relationship between arcsinh and ln, we find arcsinh(sqrt(2)a) = ln(sqrt(2)*a+sqrt(2a^2+1)), so the limit is just (1/sqrt2)*ln({sqrt(2)*a+sqrt(2a^2+1)}/{a+1}),
    which is just (3/(2*sqrt2))*ln(2). Not sure why this is off by a factor of 1/2, is it my own error, or the book’s?

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