# Solve $a, b$ in the improper integral

Find $a, b \in\mathbb{R}$ such that $$\int^\infty_1\left(\dfrac{2x^2+bx+a}{x(2x+a)}-1\right) \mathrm{d}x=1$$

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What is the asymptotic behaviour of the integrand? – Phira Nov 28 '11 at 20:58

$$I = \int_1^{\infty} \left(\frac{bx+a-ax}{x(2x+a)} \right) dx$$ $$\left(\frac{bx+a-ax}{x(2x+a)} \right) = \frac{1}{x} + \frac{b-a-2}{2x+a}$$ $$I = \int_1^{\infty} \left(\frac{1}{x} + \frac{b-a-2}{2x+a} \right) dx = \left[ \log(x) + \frac{b-a-2}{2} \log \left(x + \frac{a}{2} \right)\right]_1^{\infty}$$ Since, we get a definite answer for this, we need $b-a-2 = -2 \implies b = a$. This is so since if $b > a$, then the integral blows up to $+\infty$ and if $a > b$, the integral blows up to $- \infty$. Hence, $$I = \int_1^{\infty} \left(\frac{1}{x} + \frac{b-a-2}{2x+a} \right) dx = \left[ \log \left(\frac{x}{x+\frac{a}{2}} \right)\right]_1^{\infty} = - \log \left( \frac{1}{1+\frac{a}{2}}\right) = \log \left( 1 + \frac{a}{2} \right) = 1$$ Hence, $1 + \frac{a}{2} = e \implies a = 2 (e-1) = b$.
first simplify the integrand to $$\frac{1}{x}+\frac{b-a-2}{2x+a}$$ (divison, partial fractions). an antiderivative is $$\log(x(2x+a)^{(b-a-2)/2})$$ if $a=b$, then the limit at infinity is $-\log(2)$ (if $a\not=b$ the expression will be $\pm\infty$ at infinity), while at $1$ we have $-\log(2+a)$. Hence the integral is $\log(1+a/2)=1$ and we get $a=b=2(e-1)$
You do not explain why you suddenly impose that $a=b$. – Did Nov 28 '11 at 21:13