Definite Integral to Infinity of Exponential of -a x by One minus Cosine x over x Squared
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Theorem
- $\ds \int_0^\infty \frac {e^{-a x} \paren {1 - \cos x} } {x^2} \rd x = \arccot a - \frac a 2 \map \ln {a^2 + 1} + a \ln a$
where $a$ is a positive real number.
Proof
Set:
- $\ds \map I a = \int_0^\infty \frac {e^{-a x} \paren {1 - \cos x} } {x^2} \rd x$
for $a > 0$.
We have:
\(\ds \map {I} a\) | \(=\) | \(\ds \frac {\d^2} {\d a^2} \int_0^\infty \frac {e^{-a x} \paren {1 - \cos x} } {x^2} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac \d {\d a} \int_0^\infty \frac \partial {\partial a} \paren {\frac {e^{-a x} \paren {1 - \cos x} } {x^2} } \rd x\) | Definite Integral of Partial Derivative | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac \d {\d a} \int_0^\infty \frac {e^{-a x} \paren {1 - \cos x} } x \rd x\) | Derivative of $e^{a x}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds -\int_0^\infty \frac \partial {\partial a} \paren {\frac {e^{-a x} \paren {1 - \cos x} } x} \rd x\) | Definite Integral of Partial Derivative | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^\infty e^{-a x} \paren {1 - \cos x} \rd x\) | Derivative of $e^{a x}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^\infty e^{-a x} \rd x - \int_0^\infty e^{-a x} \cos x \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \intlimits {-\frac {e^{-a x} } a} 0 \infty - \frac a {a^2 + 1}\) | Primitive of $e^{a x}$, Definite Integral to Infinity of $e^{-a x} \cos b x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a - \frac a {a^2 + 1}\) | Exponential of Zero, Exponential Tends to Zero and Infinity |
We then have, from Primitive of $\dfrac 1 x$ and Primitive of $\dfrac x {x^2 + a^2}$:
- $\ds \map {I'} a = \ln a - \frac 1 2 \map \ln {a^2 + 1} + C$
for some constant $C \in \R$.
Note that:
\(\ds \lim_{a \mathop \to \infty} \map {I'} a\) | \(=\) | \(\ds \lim_{a \mathop \to \infty} \int_0^\infty \frac {e^{-a x} \cos x} x \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^\infty \frac 0 x \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
On the other hand:
\(\ds \lim_{a \mathop \to \infty} \map {I'} a\) | \(=\) | \(\ds \lim_{a \mathop \to \infty} \paren {\ln a - \frac 1 2 \map \ln {a^2 + 1} + C}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds C + \lim_{a \mathop \to \infty} \paren {\map \ln {\frac a {\sqrt {a^2 + 1} } } }\) | Logarithm of Power, Difference of Logarithms | |||||||||||
\(\ds \) | \(=\) | \(\ds C + \lim_{a \mathop \to \infty} \paren {\map \ln {\frac 1 {\sqrt {1 + \frac 1 {a^2} } } } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds C + \ln 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds C\) | Logarithm of 1 is 0 |
so:
- $\ds \map {I'} a = \ln a - \frac 1 2 \map \ln {a^2 + 1}$
We have:
\(\ds \map I a\) | \(=\) | \(\ds \int \map {I'} a \rd a\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \ln a \rd a - \frac 1 2 \int \map \ln {a^2 + 1} \rd a\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a \ln a - a - \frac 1 2 \paren {a \map \ln {a^ 2 + 1} - 2 a + 2 \arctan a} + C_2\) | Primitive of $\ln x$, Primitive of $\map \ln {x^2 + a^2}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds a \ln a - a - \frac 1 2 a \map \ln {a^2 + 1} + a - \arctan a + C_2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a \ln a - \frac 1 2 a \map \ln {a^2 + 1} - \arctan a + C_2\) |
for some constant $C_2 \in \R$.
Similarly we have:
\(\ds \lim_{a \mathop \to \infty} \int_0^\infty \frac {e^{-a x} \paren {1 - \cos x} } {x^2} \rd x\) | \(=\) | \(\ds \int_0^\infty \frac 0 {x^2} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
On the other hand:
\(\ds \lim_{a \mathop \to \infty} \map I a\) | \(=\) | \(\ds \lim_{a \mathop \to \infty} \map \ln {\frac {a^a} {\paren {\sqrt {a^2 + 1} }^a} } - \lim_{a \mathop \to \infty} \paren {\arctan a} + C_2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{a \mathop \to \infty} \map \ln {\frac 1 {\paren {\sqrt {1 + \frac 1 {a^2} } }^a} } - \frac \pi 2 + C_2\) | Limit of Arctangent Function | |||||||||||
\(\ds \) | \(=\) | \(\ds C_2 - \frac \pi 2\) |
We hence have:
\(\ds \map I a\) | \(=\) | \(\ds a \ln a - \frac 1 2 a \map \ln {a^2 + 1} + \paren {\frac \pi 2 - \arctan a}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a \ln a - \frac 1 2 a \map \ln {a^2 + 1} + \arccot a\) | Arccotangent Function in terms of Arctangent |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Exponential Functions: $15.89$