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Find the definite integral of:

$$\int_0^\infty \frac{\sin^4(7x)-\sin^4(5x)}{x} \ \mathrm d x$$

This question is from a Belarusian mathematical olympiad. This is from the topic of definite integrals, but I can't think of anything.

I tried using the Newton-Leibniz rule which we use generally to solve such questions. But it did not work. Please help.

Quanto
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Obsessed with integrals
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2 Answers2

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Note $\sin^4(kx )= \frac38 +\frac18\cos (4kx) -\frac12 \cos(2kx)$. Then,

\begin{align} &\int_0^\infty \frac{\sin^4(7x)-\sin^4(5x)}{x}dx\\ =&\frac18\int_0^\infty \frac{\cos(28x)-\cos(20x)}{x}dx -\frac12\int_0^\infty \frac{\cos(14x)-\cos(10x)}{x}dx\\ = &(\frac18-\frac12)\ln\frac57=\frac38\ln\frac75 \end{align}

where the result $\int_0^\infty \frac{\cos(ax)-\cos(bx )}{x}dx=\ln\frac b a$is used.

Finding $\int_0^\infty \frac{\cos(ax)-\cos(bx)}{x}dx$

Quanto
  • 97,352
3

You can use the following equality in order to compute the integral:

\begin{equation} \int\limits_{0}^{+\infty}\frac{f(t)}{t}\,\mathrm{d}t=\int\limits_{0}^{+\infty}\mathcal{L\{f(t)\}}\,\mathrm{d}s \end{equation}

Then:

\begin{equation} \int\limits_{0}^{+\infty}\frac{\sin^{4}(7t)-\sin^{4}(4t)}{t}\,\mathrm{d}t=\int\limits_{0}^{+\infty}\mathcal{L}\{\sin^{4}(7t)\}\,\mathrm{d}s-\int\limits_{0}^{+\infty}\mathcal{L}\{\sin^{4}(4t)\}\,\mathrm{d}s \end{equation}

Knowing that:

\begin{equation} \mathcal{L}\{\sin^{4}(7t)\}=\frac{57624}{s(s^{4}+980s^{2}+153664)} \end{equation}

\begin{equation} \mathcal{L}\{\sin^{4}(4t)\}=\frac{6144}{s(s^{4}+320s^{2}+16384)} \end{equation}

Thus:

\begin{equation} I=\int\limits_{0}^{+\infty}\frac{57624}{s(s^{4}+980s^{2}+153664)}\mathrm{d}s-\int\limits_{0}^{+\infty}\frac{6144}{s(s^{4}+320s^{2}+16384)}\mathrm{d}s \end{equation}

\begin{equation} I=\int\limits_{0}^{+\infty}\Biggl[\frac{57624}{s(s^{4}+980s^{2}+153664)}-\frac{6144}{s(s^{4}+320s^{2}+16384)}\Biggr]\mathrm{d}s \end{equation}

\begin{equation} I=\int\limits_{0}^{+\infty}\Biggl[\frac{57624s(s^{4}+320s^{2}+16384)-6144s(s^{4}+980s^{2}+153664)}{s^{2}(s^{4}+980s^{2}+153664)(s^{4}+320s^{2}+16384)}\Biggr]\mathrm{d}s \end{equation}

\begin{equation} I=3960\int\limits_{0}^{+\infty}\frac{s(13s^{2}+3136)}{(s^{2}+64)(s^{2}+196)(s^{2}+256)(s^{2}+784)}\mathrm{d}s \end{equation}

With $u=s^{2}$, you arrive at the following:

\begin{equation} I=1980\int\limits_{0}^{+\infty}\frac{(13u+3136)}{(u+64)(u+196)(u+256)(u+784)}\mathrm{d}u \end{equation}

This last integral is quite tedious but doable with standard methods:

\begin{equation} 1980\int\limits_{0}^{+\infty}\frac{(13u+3136)}{(u+64)(u+196)(u+256)(u+784)}\mathrm{d}u=\frac{1}{8}\ln\left(\frac{343}{64}\right) \end{equation}

Thus:

\begin{equation} \boxed{\int\limits_{0}^{+\infty}\frac{\sin^{4}(7t)-\sin^{4}(4t)}{t}\,\mathrm{d}t=\frac{1}{8}\ln\left(\frac{343}{64}\right)=\frac{3}{8}\ln\left(\frac{7}{4}\right)} \end{equation}

Luis Sierra
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