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What is the value of the following integral? $$\int_{0}^{\infty} \dfrac{1-\cos x}{x^{2}} dx$$

Mikasa
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Luis E.
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4 Answers4

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If you are aware of the rather well-known identity: $$ \forall \alpha>0,\qquad I(\alpha)=\lim_{N\to +\infty}\int_{0}^{N}\frac{\sin(\alpha x)}{x}\,dx = \frac{\pi}{2} \tag{1}$$ it is straightforward to check that: $$ J=\int_{0}^{+\infty}\frac{1-\cos(x)}{x^2}\,dx = \int_{0}^{1}I(\alpha)\,d\alpha = \color{red}{\frac{\pi}{2}}.\tag{2} $$ You may also use integration by parts to check that $J=I(1)$.

Jack D'Aurizio
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With the usual contour $\;C_{\epsilon,R}\;$ with a little "bump" around zero (canonical semicircle of radius $\;\epsilon\;$ and around the upper canonical semicircle of radius $\;R,\,\, R>>\epsilon\;$, and with the function

$$f(z)=\frac{1-e^{iz}}{z^2}\implies \text{Res}_{z=0}\,(f)= \lim_{z\to0} z\,f(z)=-ie^{i\cdot0}=-i$$implies

$$0=\lim_{\epsilon\to0,\,R\to\infty}\oint_{C_{\epsilon,R}}\,f(z)dz=\int_{-\infty}^\infty f(x)\,dx-\pi i(-i)\implies\int_0^\infty\frac{1-\cos x}{x^2}=\frac\pi2$$

by comparing real parts and noticing our function is even (and thus dividing by two)

DonAntonio
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Solution for lazy people. A special case of Ramanujan's master theorem was obtained by Glaisher in the late 19th century using non-rigorous methods. It says that if

$$f(x) = \sum_{k=0}^\infty (-1)^ka_k x^{2k}$$

then:

$$\int_0^\infty f(x) \, dx = \frac{\pi}{2} a_{-\frac{1}{2}}$$

if the integral here converges. Here one assumes a natural extrapolation of the series expansion coefficients that in the rigorous proof is made unambiguous. When this involves factorials, the prescription is to use the analytic continuation given by gamma functions. In this case, we have:

$$\frac{1-\cos(x)}{x^2} = \sum_{k=0}^\infty (-1)^k \frac{x^{2k}}{(2k+2)!}$$

therefore $a_{-\frac{1}{2}} = 1$ and the integral is thus equal to $\frac{\pi}{2}$.

Count Iblis
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You could get the answer via integration by parts for:

$$\int_0^\infty\frac{\cos(x)}{x^2} \, dx$$

via the below for suitably chosen $u(x)$ and $v(x)$;

$$\int_0^\infty v \, du = uv - \int_0^\infty u \, dv$$

Hence; $$\int_0^\infty \frac{\cos(x)}{x^2} \,dx= -\left[\cos(x) \left(\frac{-1}{x} \right) \right]_0^\infty - \int_0^\infty \frac{\sin(x)}{x}dx$$

Following Michael Hardy's comment;

$$\int_0^\infty \frac{-\sin(x)}{x} \, dx= -\left[\cos(x) \left( \frac{-1}{x} \right) \right]_0^\infty - \int_0^\infty \frac{-\cos(x)}{x^2}dx$$

We can now substitute this into the previous equation to find $\int_0^\infty \frac{\cos(x)}{x^2} \, dx$ in terms of $\cos(x)$ and $x$

For $\int_0^\infty \frac{1}{x^2} \, dx$, this is $\left[\dfrac{-1}{x}\right]_0^\infty$.

Michael Hardy
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unseen_rider
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    I don't think integration by parts will help, neither in the integral you wrote and which is not the one the OP wrote, nor the actual one...but if I am wrong I'd appreciate you describe a little more what to do. – DonAntonio Jul 18 '16 at 00:10
  • I downvoted this because it's not just an unsupported assertion but quite questionable, to say the least. $$ \int \Big(\cos x\Big) \left( \frac{dx}{x^2}\right) = \int u,dv = uv - \int v, du $$ $$ = \frac{-\cos x} x - \int\frac{-1} x(-\sin x , dx) $$ Then what do you do after that? $\qquad$ – Michael Hardy Jul 18 '16 at 02:43
  • @DonAntonio - the difference of the two integrals in my answer is the integral the OP wrote. – unseen_rider Jul 18 '16 at 09:45
  • @michaelhardy then you can do the 2nd integral by parts again, giving you a relation for $\int_0^\infty \frac{cos(x)}{x^2}dx$ in terms of $sin(x)$, $cos(x)$, and $x$ which can be solved. I will update my answer. – unseen_rider Jul 18 '16 at 09:49
  • @unseen_rider I just can't see if, so if you can show it in more detail, even if not completely detailed, perhaps that will help. – DonAntonio Jul 18 '16 at 10:01
  • @michaelhardy please read updated answer and reconsider votes and explanations. – unseen_rider Jul 18 '16 at 14:19
  • @donantonio - please revisit my answer following additional information – unseen_rider Jul 18 '16 at 14:20
  • Sorry, still can't see it. You wrote in the first line after the $;u(x),v(x);$ thing, that $$\int_0^\infty \frac{cos(x)}{x^2}dx=-[cos(x)(\frac{-1}{x}]_0^\infty - \int_0^\infty \frac{sin(x)}{x}dx$$ but I just don't succeed in seeing it: if $;u=\cos x, v'=\frac1{x^2};$ , then the first term is $;uv'=-\frac{\cos x}{x^2};$ , and you have there a double minus sign... But besides this: the expression $$\left.-\frac{\cos x}x\right|_0^\infty$$ requires taking limits, and the limit at $;x=0;$ of this doesn't exis finitely... – DonAntonio Jul 18 '16 at 16:20
  • Typo above: the first term in, of course, $;uv;$ , not $;uv';$ . – DonAntonio Jul 18 '16 at 17:05
  • @unseen_rider : Can you write $\cos(x)$ instead of $cos(x)$? $\qquad$ – Michael Hardy Jul 18 '16 at 17:56
  • @DonAntonio : You too. $\uparrow \qquad$ – Michael Hardy Jul 18 '16 at 17:56