integration of $\int_{1}^{\infty } \,\left(\frac{2x^{2}+bx\text{+}a}{x(2x+a)} -1\right) \, dx=1$

Question

i need help for this problem; Find values of a and b $$\int_{1}^{\infty} \left( \frac{2x^{2}+bx+a}{x(2x+a)} -1\right) \, dx=1$$

I very appreciate your comments and suggestions.

Answer 1

Hint: Use long division on the integrand. You should get $$\int_1^\infty \frac{-a+b-2}{a+2 x}+\frac 1x+1-1$$ $$=\int_1^\infty\frac{-a+b-2}{a+2 x}+\int_1^\infty\frac 1x$$ $$=\lim_{t\to\infty}(-a+b-2)\int_1^t \frac{1}{a+2x}+\int_1^t \frac 1x$$

Can you take it from there?

Answer 2

For \begin{align} I = \int_{1}^{\infty} \left( \frac{2 x^2 + b x + a}{x(2x+a)} - 1 \right) \, dx = 1 \end{align} then it is seen that \begin{align} I &= \lim_{p \to \infty} \, \int_{1}^{p} \left( \frac{b-a}{2x+a} + \frac{a}{x(2x+a)} \right) \, dx = \lim_{p \to \infty} \, \int_{1}^{p} \left( \frac{b-a-2}{2x+a} + \frac{1}{x} \right) \, dx \\ &= \lim_{p \to \infty} \, \left[ \frac{b-a-2}{2} \, \ln(2x+a) + \ln(x) \right]_{1}^{p} \\ &= \lim_{p \to \infty} \ln\left[ p \, \left(\frac{2p+a}{2+a} \right)^{\frac{b-a-2}{2}} \right] = \ln(e) = 1 \end{align} for which \begin{align} \lim_{p \to \infty} \left\{ p \, \left(\frac{2p+a}{2+a} \right)^{\frac{b-a-2}{2}} \right\} = e. \end{align}

The limit is evaluated as follows: \begin{align} e &= \lim_{p \to \infty} \left\{ p \left(\frac{2p+a}{2+a}\right)^{\frac{b-a-2}{2}} \right\} \\ &= \left(\frac{2}{a+2}\right)^{\frac{b-a-2}{2}} \, \lim_{p \to \infty} \left\{ p^{\frac{b-a}{2}} \, \left( 1 + \frac{a}{2p}\right)^{\frac{b-a-2}{2}} \right\} = \left(\frac{2}{a+2}\right)^{\frac{b-a-2}{2}} \, \lim_{p \to \infty} \left\{ p^{\frac{b-a}{2}} \right\} \end{align} If $b=a$ then the limit is reduced to $1$. This yields \begin{align} e = \frac{a + 2}{2} \end{align} leading to $a = b = 2(e - 1)$.