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I have been having trouble understanding Fourier series and analysis in one of my classes. This is one problem from the text and we have to show that this is true. I have done other problems related to this one, but they do not help me. I tried to solve this manually and it has come to naught. As there is already the answer, I would like an explanation so that I may understand this material better. Thanks for all the help.

$$\int^{\infty}_{0}{(1-\cos x)\over{x^{2}}}dx = \frac\pi{2}$$

This is the integral for reference.

Anastasiya-Romanova 秀
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Chris
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3 Answers3

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Since $1-\cos x=2\sin^2(x/2)$ we have $$ I:=\int^{\infty}_{0}\frac{(1-\cos x)}{x^2}dx =\int_{0}^{\infty}\frac{4\sin^2(x/2)}{x^2}dx $$ After substitution $t=x/2$ we get $$ I=\int_{0}^{\infty}\frac{2\sin^2 t}{t^2}dt $$ Since the function $\frac{2\sin^2 t}{t^2}$ is even, we can say $$ I=\frac{1}{2}\int_{-\infty}^{\infty}\frac{2\sin^2 t}{t^2}dt =\int_{-\infty}^{\infty}\frac{\sin^2 t}{t^2}dt $$ Now see this answer

Community
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Norbert
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Another way to obtain the answer is through the Dirichlet integral. Notice that:

$$\int_{0}^{\infty}\frac{1-\cos(x)}{x^{2}}dx=\int_{0}^{\infty}\frac{\sin(x)}{x}dx$$

after an application of integration by parts where I integrated $\frac{1}{x^{2}}$ and differentiated $1-\cos(x)$. Then look at Dirichlet integral.

Anastasiya-Romanova 秀
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user71352
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  • This helps understand why I do not know how this works, but I have to use Fourier cosine transforms to show this relation. I have solved that in a previous homework problem, but to reach this answer from a cosine transform is still blurry; considering Laplace transforms are the easiest way, and my professor does not like them so he didn't teach us. – Chris Apr 23 '14 at 12:44
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We have the following property for Laplace transforms: $$\mathcal{L} \left\{ \frac{f(t)}{t} \right\}(s)=\int_s^\infty F(p)dp. $$ Replacing $f(t)$ with $\frac{f(t)}{t}$ gives a slight generalization:

$$\mathcal{L} \left\{ \frac{f(t)}{t^2} \right\}(s)=\int_s^\infty \mathcal{L} \left\{ \frac{f(t)}{t} \right\}(p)dp=\int_s^\infty \int_p^\infty F(q)dqdp. $$

Finally, using $f(t)=1-\cos t$ in the above and taking $s=0$ gives

$$\int_0^\infty \frac{1-\cos t}{t^2} e^{-0t}dt=\int_0^\infty \int_p^\infty \left( \frac{1}{q}-\frac{q}{q^2+1} \right)dq dp=\int_0^\infty \log q-\frac{1}{2}\log (q^2+1) \big]_{q=p}^{q=\infty} dp =\int_0^\infty \left( \frac{1}{2} \log(p^2+1)-\log p \right) dp $$ and the last integral can evaulated to $\frac{\pi}{2}$ via integration by parts.

user1337
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  • Thank you for replying, but I think I should have included that Fourier cosine transforms are supposed to be our catalyst for understanding this integral. We did not learn Laplace transforms as my professor thinks they are the easy way around Fourier analysis. How can Fourier cosine transforms be used to interpret this? – Chris Apr 23 '14 at 12:47