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

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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

\(\ds \int_0^\infty e^{-a x} \sin b x \rd x\) \(=\) \(\ds \intlimits {\frac {e^{-a x} \paren {-a \sin b x - b \cos b x} } {a^2 + b^2} } 0 \infty\) Primitive of $e^{a x} \sin b x$
\(\ds \) \(=\) \(\ds -\lim_{x \mathop \to \infty} \paren {\frac {e^{-a x} \paren {a \sin b x + b \cos b x} } {a^2 + b^2} } + \frac {e^0 \paren {a \sin 0 + b \cos 0} } {a^2 + b^2}\)
\(\ds \) \(=\) \(\ds -\lim_{x \mathop \to \infty} \paren {\frac {e^{-a x} \paren {a \sin b x + b \cos b x} } {a^2 + b^2} } +\frac b {a^2 + b^2}\) Exponential of Zero, Sine of Zero is Zero, Cosine of Zero is One

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:

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

$\blacksquare$


Sources

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