How do we prove that $$\int_{1}^{\infty} \frac{\sin^4(\log x)}{x^2 \log x} \mathrm{d}x=\dfrac{\log\left(\dfrac{625}{17}\right)}{16}$$

I tried substitutions like $\log x=\arcsin t$, but it doesn't seem to work out. Please help me out. Thank you.

  • $\begingroup$ Let $\log x=t$. $\endgroup$ – Ali Caglayan Nov 2 '14 at 18:16
  • $\begingroup$ @Alizter then what should I do? $\endgroup$ – pkwssis Nov 2 '14 at 18:23
  • $\begingroup$ Wbhat have you tried? Please edit the question to include additional context, including your attempts and also the context in which you encountered the integral. $\endgroup$ – Carl Mummert Nov 2 '14 at 18:25
  • $\begingroup$ Do you know complex integration? $\endgroup$ – Timbuc Nov 2 '14 at 18:26
  • 1
    $\begingroup$ @Pkwssis: yes, there is no hurry, of course. I apologize if I made it sound urgent. $\endgroup$ – Carl Mummert Nov 2 '14 at 18:33

The integral screams for a sub $x=e^u$; the result is

$$\int_0^{\infty} du \, e^{-u} \frac{\sin^4{u}}{u} $$

This is very computable by introducing a parameter and differentiating under the integral. In this case, consider

$$F(k) = \int_0^{\infty} du \, e^{-k u} \frac{\sin^4{u}}{u} $$

$$F'(k) = -\int_0^{\infty} du \, e^{-k u} \sin^4{u} $$

$F'(k)$ is relatively easy to compute using the fact that $\sin^4{u} = \frac{3}{8}-\frac12 \cos{2 u} + \frac18 \cos{4 u}$, and that

$$\int_0^{\infty} du \, e^{-k u} \cos{m u} = \frac{k}{k^2+m^2}$$


$$F'(k) = -\frac{3}{8 k} + \frac12 \frac{k}{k^2+4} - \frac18 \frac{k}{k^2+16} $$


$$F(k) = -\frac1{16} \log{\left [ \frac{k^6 (k^2+16)}{(k^2+4)^4} \right ]} + C$$

To evaluate $C$, we must consider $\lim_{k \to \infty} F(k)$ because $F(0)$ represents a non convergent integral. Because the limit is zero, we must have $C=0$. The integral we seek is then

$$F(1) = \frac1{16} \log{\frac{625}{17}}$$

  • $\begingroup$ We could also use the laplace transform to evaluate the integral.after the substitution, right? $\endgroup$ – pkwssis Nov 3 '14 at 0:50
  • $\begingroup$ @Pkwssis: Well, the Laplace transform is just a name for the integral. It doesn't really offer a method of evaluating the integral, unless you count looking up in tables. $\endgroup$ – Ron Gordon Nov 3 '14 at 0:53

I propose the following approach:

  • Let $t=\dfrac1x$
  • Let $u=\ln t$.
  • Use the power-reduction formula for $\sin^4x$.
  • Employ Euler's formula in conjunction with the linear properties of the integral.
  • Express your new integral(s) in terms of $I(k)=\displaystyle\int_0^\infty e^{-kx}~dx$. You'll have $\displaystyle\int I(k)~dk$, where

$\qquad I(k)$ will turn out to be a rational function in k. Solve it by using the usual methods.


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