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

Find a value for the constant so that the improper integral

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

First, we have

For this to converge we must have the coefficient of the in the numerator going to 0 as (otherwise the integral will converge by limit comparison with ). Hence,

Now, we need to evaluate the integral. Incomplete.

1. S says:

You already solved earlier (maybe in chapter 8): \int \frac{dx}{\sqrt{1+x^2}}, by substituting t = x + \sqrt{1 + x^2}. This integral can similarly be solved.

2. 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?