A definite integral from 0 to infinity

Relevant wiki: Differentiation Under the Integral Sign

Define the function $F(s) = \int_{0}^{\infty} \frac{\sin 3x \sin 5x}{x} e^{-sx} dx$ . Notice that the integral we want to find is $F(0)$ which we may compute if we are able to find another expression for $F$ . For this, compute $F'(s) = \int_{0}^{\infty} \frac{\sin 3x \sin 5x}{x} e^{-sx} (-x) dx = -\int_{0}^{\infty} \sin 3x \sin 5x e^{-sx} dx$ . To be able to compute this integral first apply the identity $\sin 3x \sin 5x = \frac{1}{2} \left( \cos 2x - \cos 8x \right)$ so we have $F'(s) = \frac{1}{2} \int_{0}^{\infty} (\cos 8x - \cos 2x) e^{-sx} dx$ . These are now standard integrals the reader should be able to compute, but if one knows about Laplace transforms then this can be written as $F'(s) = \frac{1}{2} \left( \mathscr{L} \left( \cos 8x \right) - \mathscr{L} \left( \cos 2x \right) \right)$ and using the known formulas one can find that this is equal to $F'(s) = \frac{1}{2} \left( \frac{s}{s^2 + 8^2} - \frac{s}{s^2 + 2^2} \right)$

To find $F(s)$ we want to integrate this with respect to $s$ which we may do by first computing $\int \frac{s}{s^2 + a^2} ds$ . Let $u = s^2 + a^2$ so $du = 2s ds \implies sds = \frac{du}{2}$ so the integral is equal to $\frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \log u = \frac{1}{2} \log(s^2 + a^2)$ . Applying this result we get that:

$F(s) = \frac{1}{4} \left( \log(s^2 + 8^2) - \log(s^2 + 2^2) \right) + C$ where $C$ is some constant. To find what $C$ is we just need a little more information about $F(s)$ . Notice that according to the current expression for $F(s), \lim_{s \to \infty} F(s) = C$ . But we can compute this limit using the original definition:

$0 \leq \left| \int_{0}^{\infty} \frac{\sin 3x \sin 5x}{x} e^{-sx} dx \right | \leq \int_{0}^{\infty} \left| \frac{\sin 3x \sin 5x}{x} e^{-sx} \right| dx \leq \int_{0}^{\infty} 3e^{-sx}dx = \frac{3}{s} \to 0$ as $s \to \infty$ .

According to the previous chain of inequalities and the Squeeze Theorem, this limit is 0 which implies $C = 0$ so we can finally say that $F(s) = \frac{1}{4} \left( \log(s^2 + 8^2) - \log(s^2 + 2^2) \right)$ and

$F(0) = \frac{1}{4} \log \left(\frac{8^2}{2^2} \right) = \frac{1}{4} \log 16 = \frac{1}{4} \log(2^4) = \log 2$ .

Let: $I = \large \int \limits_{0}^{\infty} \frac {\sin 3x \sin 5x}{x} \,dx\ = \large \frac{1}{2} \large \int\limits_{0}^{\infty} \frac {\cos 2x - \cos 8x}{x}\,dx\ =\large \frac{1}{2} \large \int\limits_{0}^{\infty} \frac {f(ax) - f(bx)}{x} \,dx\ \,$ , where $f(x) = \cos x$ , $a = 2$ and $b = 8$ .

(Now we use the results established by Frullani for integrals of the form: $\large \int\limits_{0}^{\infty} \frac {f(ax) - f(bx)}{x} \,dx\$ , where $f$ is a continuous function in $\large \mathbb R^+$ and $\large a$ , $\large b$ $\large \in \mathbb R^+$ .

See here for Frullani Theorem . Refer conditions $C1$ and $C3$ . They are applicable in this case as $f(x) = \cos x$ satisfies them both).

Thus, by using the result when conditions $C1$ and $C3$ applicable:

$\large {I = \frac {1}{2}\large \left( \lim_{x\to0} f(x) \right) \log \frac{b}{a} = \large \frac {1}{2} \log 4 = \log 2}$