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$$ Y(a) = \int_{0}^{\infty} \frac{e^{-ax}}{x^2+4} dx $$

a) Find the values of $a$ for which $Y(a)$ is well defined.

b) $Y$ checks on $(0, \infty)$ a non-homogeneous differential equation of degree two with constant coefficients.

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    $\begingroup$ What number is $a$ here? $\endgroup$ – Dr. Sonnhard Graubner Jul 5 at 15:14
  • $\begingroup$ Any real number. $\endgroup$ – SADBOYS Jul 5 at 15:17
  • $\begingroup$ It must be $$\Re(a)>0$$ $\endgroup$ – Dr. Sonnhard Graubner Jul 5 at 15:23
  • $\begingroup$ Can you explain your second question? I'm not sure what you mean. $\endgroup$ – automaticallyGenerated Jul 5 at 15:23
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    $\begingroup$ @DrSonnhardGraubner $$\Re(a) \ge 0$$ to be exact. $\endgroup$ – automaticallyGenerated Jul 5 at 15:26
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a) The integral exists if and only if $a \geq 0$. For $a \geq 0$, compare the integrand to $\frac{1}{x^2+4}$ to see that the integral converges. If $a < 0$, then there is an $x_0$ such that the integrand is positive and increases for all $x > x_0$, which shows that the integral can not be finite.

b) Differentiating twice under the integral sign, we obtain for $a > 0$ $$Y''(a) = \int_{0}^{\infty} \frac{x^2}{x^2 + 4} e^{-ax}~ \mathrm dx.$$ This yields $$Y''(a) + 4 Y(a) = \int_{0}^{\infty} e^{-ax}~\mathrm dx = \frac{1}{a}, \quad a > 0.$$ which is the desired differential equation.

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