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How to calculate

$$\lim_{y\rightarrow +\infty}\left( \int_0^y \frac {\sin^4(x)}{x} \mathrm{d}x - \frac38 \ln y\right).$$

In my view,there is $$\int_0^y \frac{\sin^4 x}{x} \mathrm{d} x =\int_0^y\frac{ \cos 4 x - 4 \cos 2 x + 3}{8 x}\mathrm{d} x.$$ I try using the Taylor expansion of cosine functions and logarithmic functions to calculate this limit，but this way seem to be not work. I think there is another way to calculate it, whereas I haven't figured it out yet.

metamorphy
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Mr.He
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  • "Can we calculate this limit using the Taylor expansion of cosine functions and logarithmic functions?" Why don't you give it a go ? This is your problem, not ours. We can help you where you're stuck, not where you're too lazy to try. [Not trying to be rude, but I think you have shown little effort in your question...] – Adam Rubinson May 05 '23 at 05:52
  • I mean that I have tried this way,but get nothing@AdamRubinson – Mr.He May 05 '23 at 06:17
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    Ok, then you should write something like, "I have tried using Taylor expansions... but didn't get anywhere". – Adam Rubinson May 05 '23 at 06:36

1 Answers1

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Writing $\int_0^y=\int_0^1+\int_1^y$ and $\ln y=\int_1^y\frac{dx}{x}$, we see that the limit equals $$\int_0^1\frac{\cos 4x-4\cos 2x+3}{8x}\,dx+\int_1^\infty\frac{\cos 4x-4\cos 2x}{8x}\,dx=\frac{f(2)}{2}-\frac{f(4)}{8}$$ where, for $a>0$, $$f(a)=\int_0^1\frac{1-\cos ax}{x}\,dx-\int_1^\infty\frac{\cos ax}{x}\,dx=\int_0^a\frac{1-\cos t}{t}\,dt-\int_a^\infty\frac{\cos t}{t}\,dt.$$

Writing $\int_0^a=\int_0^1+\int_1^a$ and $\int_a^\infty=\int_1^\infty-\int_1^a$, we see that $f(a)=f(1)+\ln a$.

Finally, it is known that $f(1)=\gamma$ equals the Euler–Mascheroni constant; a possible way to show this is to use the integral with $e^{-t}$ in place of $\cos t$, and the fact that $\int_0^\infty\frac{\cos t-e^{-t}}{t}\,dt=0$.

Put together, we get the answer $$\lim_{y\to+\infty}\left(\int_0^y\frac{\sin^4 x}{x}\,dx-\frac38\ln y\right)=\frac{3\gamma+2\ln2}8.$$

metamorphy
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