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

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

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.

• What number is $a$ here? – Dr. Sonnhard Graubner Jul 5 at 15:14
• Any real number. – SADBOYS Jul 5 at 15:17
• It must be $$\Re(a)>0$$ – Dr. Sonnhard Graubner Jul 5 at 15:23
• Can you explain your second question? I'm not sure what you mean. – automaticallyGenerated Jul 5 at 15:23
• @DrSonnhardGraubner $$\Re(a) \ge 0$$ to be exact. – automaticallyGenerated Jul 5 at 15:26

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.