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Find values of constants a and b such that the given improper integral converges

Find values for real numbers a and b so that the following improper integral equation holds:

    \[ \int_1^{\infty} \left( \frac{2x^2 + bx + a}{x(2x+a)} - 1 \right) \, dx = 1. \]


First, we have

    \begin{align*}  \int_1^{\infty} \left( \frac{2x^2 + bx + a}{x(2x+a)} - 1 \right) \, dx &= \int_1^{\infty} \frac{2x^2 + bx + a - 2x^2 - ax}{2x^2+ax} \, dx \\[9pt]  &= \int_1^{\infty} \frac{(b-a)x + a}{2x^2 + ax} \, dx. \end{align*}

In order for this to converge we must have the coefficient of x in the numerator equal to 0; hence, we must have a = b. Now, to solve the integral equation we have

    \begin{align*}  \int_1^{\infty} \frac{a}{2x^2 + ax} \,dx &= \int_1^{\infty} \frac{a}{x(2x+a)} \, dx \\[9pt]  &= \int_1^{\infty} \left( \frac{1}{x} - \frac{2}{2x+a} \right) \, dx &(\text{partial fractions})\\[9pt]  &= \lim_{c \to \infty} \int_1^c \left( \frac{1}{x} - \frac{2}{2x+a} \right) \, dx \\[9pt]  &= \lim_{c \to \infty} \left( \log |x| - \log |2x+a| \right)\Bigr \rvert_1^c \\[9pt]  &= \lim_{c \to \infty} \left( \log \left| \frac{x}{2x+a} \right| \right) \Bigr \rvert_1^c \\[9pt]  &= \lim_{c \to \infty} \left( \log \left( \frac{c}{2c+a} \right) - \log \left( \frac{1}{2+a} \right) \right) \\[9pt]  &= \log \frac{1}{2} - \log \frac{1}{2+a} \\[9pt]  &= \log \frac{2+a}{2}. \end{align*}

But, by assumption this integral equals 1, so we have

    \begin{align*}  \log \frac{2+a}{2} = 1 && \implies && \frac{2+a}{2} &= e \\[9pt]  && \implies && 2+a &= 2e \\[9pt]  && \implies && a & =2e - 2. \end{align*}

Since b = a we then have

    \[ a = b = 2e - 2. \]

One comment

  1. Rafael Deiga says:

    In the second integral, from fifth to sixth row, why did you remove the module? If you keep the module until the end, other solution gonna appear. The other solution is a= -2e-2.

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