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Umaiza Hutton

Answered question

2022-02-28

How can I integrate the following:

\frac{1}{b^{2}} \int_{0}^{\infty} z^{- 2} exp (- a z) \sin^{2} (b z) d z

for

a, b > 0

Spandri3uv

Beginner2022-03-01Added 7 answers

Let

I (a)

denote the integral. Consider

I^{″} (a) = \frac{1}{b^{2}} \int_{0}^{\infty} exp (- a z) \sin^{2} (b z) d z

= \frac{1}{2 b^{2}} \int_{0}^{\infty} exp (- a z) (1 - \cos (2 b z)) d z

= \frac{1}{2 b^{2}} R \int_{0}^{\infty} (exp (- a z) - exp (- (a + 2 i b) z)) d z

= \frac{1}{2 b^{2}} R (\frac{1}{a} - \frac{1}{a + 2 i b}) = \frac{2}{a} \frac{1}{a^{2} + 4 b^{2}}

Now integrate back, using

I (\infty) = 0

and

I^{'} (\infty) = 0

I^{'} (a) = \int_{a}^{\infty} \frac{2}{t} \frac{1}{t^{2} + 4 b^{2}} d t = \frac{1}{4 b^{2}} \log (1 + \frac{4 b^{2}}{a^{2}})

I (a) = \int_{a}^{\infty} I^{'} (t) d t = \frac{1}{b} \arctan (\frac{2 b}{a}) - \frac{a}{4 b^{2}} \log (1 + \frac{4 b^{2}}{a^{2}})

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