Definite Integral to Infinity of Exponential of -a x by Sine of b x

Theorem

 * $\ds \int_0^\infty e^{-a x} \sin b x \rd x = \frac b {a^2 + b^2}$

where $a$ and $b$ are real numbers with $a > 0$.

Proof
Note that we have, by Linear Combination of Sine and Cosine:


 * $\ds 0 \le \size {\frac {e^{-a x} \paren {a \sin b x + b \cos b x} } {a^2 + b^2} } \le \frac {e^{-a x} \sqrt {a^2 + b^2} } {a^2 + b^2} = \frac {e^{-a x} } {\sqrt {a^2 + b^2} }$

By Exponential Tends to Zero and Infinity:


 * $\ds \lim_{x \mathop \to \infty} \paren {\frac {e^{-a x} } {\sqrt {a^2 + b^2} } } = 0$

So by the Squeeze Theorem:


 * $\ds \lim_{x \mathop \to \infty} \paren {\frac {e^{-a x} \paren {a \sin b x + b \cos b x} } {a^2 + b^2} } = 0$

So:


 * $\ds \int_0^\infty e^{-a x} \sin b x \rd x = \frac b {a^2 + b^2}$