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So we have the following improper integral and we have to prove that it converges(if it converges then find its value). $$\int_0^\infty \frac{\sin x(1-\cos x)}{x^2}dx$$

I tried the feynman approach of letting $f(a)$ as: $$f(a)=\int_0^\infty \frac{\sin(ax)(1-\cos(ax))}{x^2}dx$$ $$f(a)=\int_0^\infty \frac{\sin(ax)-\sin(2ax)/2}{x^2}dx$$ Differentiating twice, wrt $a$: $$f''(a)=-\int_0^\infty \sin(ax)-2\sin(2ax)dx$$ But then I run into a trap because of limits, I have to let $\lim_{x\rightarrow \infty}(\cos(ax))=\alpha;\,\,\,[-1<\alpha<1]$, but it is fruitless, and I can't solve it further. So I need help after this stage or could anyone suggest a better method? Thanks in advance.

Shubham Kumar
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Kshitij Kumar
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    It might help that by the double-angle identities, $\sin(x)[1-\cos(x)] = 2\sin(x)\sin^2(x/2) = 4 \sin^3(x/2)\cos(x/2)$. – eyeballfrog Sep 18 '23 at 17:15
  • Integrate by parts and see here – user170231 Sep 18 '23 at 17:26
  • I think after using the double angle identities, series expansion and Ramanujan's master theorem works. – bob Sep 18 '23 at 17:32
  • Substituting $x\rightarrow 2u$ lets you write it as $\int_0^\infty \sin(2u)[\sin(u)/u]^2 dx$, which is so close to being the Fourier transform of the triangle function. – eyeballfrog Sep 18 '23 at 17:34
  • Your integral is a special case of the one in https://math.stackexchange.com/questions/1643989/evaluating-int-0-infty-frac-sinxt1-cosatt2-dt?rq=1. – Gonçalo Sep 18 '23 at 19:06

3 Answers3

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We can rewrite as $$I = \int_{0}^{\infty}\frac{\sin{(x)}(1 - \cos{(x)})}{x^{2}} \ dx = \int_{0}^{\infty}\left(\frac{\sin{x}}{x} - \frac{\sin{2x}}{2x}\right)\frac{1}{x}\ dx$$ This is of the form of a Frullani Integral where $f(x) = \frac{\sin{x}}{x}$ for $x > 0$ and $f(x) = 1$ for $x = 0$. We have that $f(0) = 1$ and $f(\infty) = 0$. Then, $$I = (f(\infty) - f(0))\ln\frac{1}{2} = (0 - 1)\ln\frac{1}{2} = \ln{2}$$

conan
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You can use the following property $$ \int_0^\infty \mathcal{L}\{f(t)\}(x) g(x)\;dx=\int_0^\infty \mathcal{L}\{g(t)\}(x) f(x)\;dx $$ to get the integral. In fact, noting $$ \mathcal{L}\{t\}(x)=\frac1{x^2}$$ one has \begin{eqnarray} I &=& \int_{0}^{\infty}\frac{\sin{(x)}(1 - \cos{(x)})}{x^{2}} \ dx \\ &=& \int_{0}^{\infty}\sin{(x)}(1 - \cos{(x)})\mathcal{L}\{t\}(x)\; dx\\ &=& \int_{0}^{\infty}x\mathcal{L}\{\sin{(t)}(1 - \cos{(t)})\}(x)\; dx\\ &=&\int_{0}^{\infty}\frac{3x}{(x^2+1)(x^2+4)}\; dx\\ &=&\frac12\int_{0}^{\infty}\frac{3}{(x+1)(x+4)}\; dx\\ &=&\frac12\ln\bigg(\frac{x+1}{x+4}\bigg)\bigg|_0^\infty\\ &=&\ln2. \end{eqnarray}

xpaul
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Integration by parts gives $$ \begin{aligned} I & =-\int_0^{\infty} \sin x(1-\cos x) d\left(\frac{1}{x}\right) \\ & =- \underbrace{ \left[\frac{\sin x(1-\cos x)}{x}\right]_0^{\infty}}_{=0} +\int_0^{\infty} \frac{\cos x-\cos (2 x)}{x}dx \end{aligned} $$ Considering the parameterized integral $$ I(a)=\int_0^{\infty} \frac{\cos x-\cos (2 x)}{x} e^{-a x} d x $$ Differentiating w.r.t. $a$ yields $$ \begin{aligned} I^{\prime}(a) & =-\int_0^{\infty} \cos x e^{-a x} d x+\int_0^{\infty} \cos (2 x) e^{-a x} d x \\ & =-\frac{a}{a^2+1}+\frac{a}{a^2+4} \end{aligned} $$ Integrating $I’(a) $ from $a=0$ to $\infty$ gives $$ I(\infty)-I(0)=\int_0^{\infty}\left(-\frac{a}{a^2+1}+\frac{a}{a^2+4}\right) d a=-\ln 2 $$ Hence $$I=I(0)=\ln 2$$

Lai
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