How to integrate $I = \int_{0}^{\infty} \frac{e^{-x}\sin^4(x)}{x^{2}}\,dx$

Question

Question: $$I = \int_{0}^{\infty} \frac{e^{-x}\sin^{4}(x)}{x^{2}}dx$$ My Approach, I tried using feynman technique to get rid of the denominator $$I(a) = \int_{0}^{\infty} \frac{e^{-ax}\sin^{4}(x)}{x^{2}}dx$$ From this we have our boundary condition $I(0) = \frac{\pi}{4}, \hspace{1mm}I(\infty) = 0 \hspace{1.5mm} and \hspace{1.5mm}I'(\infty)=0$ $$I'(a) = \int_{0}^{\infty} \frac{-e^{-ax}\sin^{4}(x)}{x}dx$$ $$I''(a) = \int_{0}^{\infty} e^{-ax}\sin^{4}(x)dx$$ So integrating this we get: $$I''(a) = \frac{24}{a^{5}+20a^{3}+64a}$$ So integrating this twice ad applying boundary conditions we get $$I(a) = \int \int \left(\frac{24}{a^{5}+20a^{3}+64a}da \right)da = \frac{a}{16}\ln{\left(\frac{a^{6}(a^{2}+16)}{(a^{2}+4)^{4}}\right)} + 8\arctan\left(\frac{a}{4}\right)-16\arctan\left(\frac{a}{2}\right)+\frac{\pi}{4}$$ Since $I(1)$ is are target integral we get $$\int_{0}^{\infty} \frac{e^{-x}\sin^{4}(x)}{x^{2}}dx = \frac{-1}{16}\ln{\left(\frac{5^{4}}{17}\right)} + 8\arctan\left(\frac{1}{4}\right)-16\arctan\left(\frac{1}{2}\right)+\frac{\pi}{4} \approx -4.89841$$ But the actual answer seems to be diffrent, it is $$\int_{0}^{\infty} \frac{e^{-x}\sin^{4}(x)}{x^{2}}dx \approx 0.218965$$ So, My question is where have I gone wrong or is their another way to solve this.

Answer 1

$$I''(a) = \frac{24}{a^{5}+20a^{3}+64a}$$ $$I'(a) = -\frac{1}{4} \log \left(a^2+4\right)+\frac{1}{16} \log \left(a^2+16\right)+\frac{3 \log (a)}{8}+c_1$$ $$I(a) =-\frac{1}{4} a \log \left(a^2+4\right)+\frac{1}{16} a \log \left(a^2+16\right)+\frac{3}{8} a \log (a)+$$ $$\frac{1}{2} \tan ^{-1}\left(\frac{a}{4}\right)-\tan ^{-1}\left(\frac{a}{2}\right)+a c_1+c_2$$

Using $I(0)=\frac \pi 4$ makes $c_2=\frac \pi 4$.

Making $a=1$, we then have $$c_1+\frac{\pi }{4}-\frac{1}{16} \log \left(\frac{625}{17}\right)-\frac{1}{4} \tan ^{-1}\left(\frac{416}{87}\right)$$

Answer 2

A slightly brief approach, just too long for a comment, notice $$ \begin{aligned} \int_{0}^{\infty} \frac{e^{-x}\sin^4x}{x^2} \,\mathrm{d}x & = \int_{0}^{\infty} \frac{e^{-x}\sin^2x(1-\cos^2x)}{x^2} \,\mathrm{d}x\\ & = \int_{0}^{\infty} \frac{e^{-x}\sin^2x}{x^2} \,\mathrm{d}x - \frac1{2} \int_{0}^{\infty} \frac{e^{-x/2}\sin^2x}{x^2} \,\mathrm{d}x \end{aligned} $$ notice that our separating of improper integral is valid here, denote $$ J(a) = a\int_{0}^{\infty} \frac{e^{-ax}\sin^2x}{x^2} \,\mathrm{d}x = aK(a) $$ By the same argument $K(0)=\pi/2$, $K'(\infty)=0$ $$ K''(a) = \int_{0}^{\infty} e^{-ax}\sin^2x \,\mathrm{d}x = \frac2{a(a^2+4)} $$ $$ K'(a) = -\int_{a}^{\infty} \frac{2\,\mathrm{d}u}{u(u^2+4)} = \frac1{4}\ln\left(\frac{a^2}{a^2+4}\right) $$ $$ K(a) = K(0) + \int_{0}^{a} \frac1{4}\ln\left(\frac{u^2}{u^2+4}\right) \mathrm{d}u = \frac{\pi}{2} - \arctan\left(\frac{a}{2}\right) + \frac{a}{4}\ln\left(\frac{a^2}{a^2+4}\right) $$ $$ J(a) = \frac{\pi a}{2} - a\arctan\left(\frac{a}{2}\right) + \frac{a^2}{4}\ln\left(\frac{a^2}{a^2+4}\right) $$ What we need is $I(1)=J(1)-J(1/2)$ which will gives you same answer. Almost the same approach, yet this one maybe more clearly shows the structure of $J(a)$.

Answer 3

Using the trig identity $$\sin^{4}(x) = \frac{3-4 \cos(2x)+\cos (4x)}{8}, $$ we have $$ \begin{align} \int_{0}^{\infty} \frac{e^{-x} \sin^{4}(x)}{x^{2}} \, \mathrm dx &= \int_{0}^{\infty} \frac{e^{-x}}{x^{2}} \frac{3-4 \cos(2x) + \cos(4x)}{8} \, \mathrm dx \\ &= \frac{1}{8}\lim_{s \to -1} \int_{0}^{\infty} e^{-x} \left(3 - 4 \cos(2x) + \cos(4x) \right) \, x^{s-1} \, \mathrm dx \\ &= \frac{1}{8} \lim_{s \to -1} I(s).\end{align}$$

First assume that $\Re(s) >0$.

Using the integral formula $$\int_{0}^{\infty} e^{-x} \cos(ax) x^{s-1} \, \mathrm dx = \frac{\Gamma(s) \cos (s \arctan a) }{(a^{2}+1)^{s/2}} , \quad (\Re(s) >0, \ a \in \mathbb{R}), $$ we have

$$ \begin{align} I(s) &= 3\int_{0}^{\infty} e^{-x} x^{s-1} \, \mathrm dx -4 \int_{0}^{\infty}e^{-x} \cos(2x) x^{s-1} \, \mathrm dx + \int_{0}^{\infty} e^{-x} \cos(4x) x^{s-1} \, \mathrm dx \\ &=\underbrace{\Gamma(s) \left(3 -4 \, \frac{ \cos(s \arctan 2)}{5^{s/2}} + \frac{\cos (s \arctan 4)}{17^{s/2}}\right)}_{f(s)}. \end{align}$$

The Mellin transform of a function defines an analytic function where the integral converges absolutely.

And $f(s)$ has removable singularities at $s=0, -1, -2$, and $-3$.

If we assign $f(s)$ it's limit values at these points, then $I(s) = f(s)$ for $\Re(s) >-4$ by analytic continuation.

Therefore, using L'Hôpital's rule, we have

$ \begin{align} \lim_{s \to -1} I(s) &= \lim_{s \to -1} \Gamma(s) \left(3-\frac{4 \cos(s \arctan 2)}{5^{s/2}} + \frac{\cos(s \arctan 4)}{17^{s/2}}\right) \\ &= \lim_{s \to -1} \frac{3-\frac{4 \cos(s \arctan 2)}{5^{s/2}} + \frac{\cos(s \arctan 4)}{17^{s/2}}}{\frac{1}{\Gamma(s)}} \\ &=\small\lim_{s \to -1} \frac{-4 \frac{-\sin(s \arctan 2)\arctan(2) 5^{s/2}- \cos(s \arctan 2)\frac{5^{s/2}}{2} \ln(5)}{5^{s}}+ \frac{-\sin(s \arctan 4)\arctan(4) 17^{s/2}- \cos(s \arctan 4)\frac{17^{s/2}}{2} \ln(17)}{17^{s}}}{- \frac{\psi(s)}{\Gamma(s)}} \\ &\overset{\diamondsuit}{=} \frac{-20 \left(\frac{2}{\sqrt{5}} \, \arctan(2) \frac{1}{\sqrt{5}}-\frac{1}{\sqrt{5}} \frac{1}{2 \sqrt{5}} \, \ln(5) \right)+ 17 \left( \frac{4}{\sqrt{17}} \, \arctan(4) \frac{1}{\sqrt{17}}-\frac{1}{\sqrt{17}} \frac{1}{2\sqrt{17}} \ln(17) \right)}{-1} \\ &= 8 \arctan(2) -2 \ln(5) - 4 \arctan(4) + \frac{\ln(17)}{2}, \end{align}$

and $$\int_{0}^{\infty} \frac{e^{-x}\sin^{4}(x)}{x^{2}} \, \mathrm dx = \frac{1}{8} \left(8 \arctan(2) -2 \ln(5) - 4 \arctan(4) + \frac{\ln(17)}{2} \right) = 0.218956 \ldots $$

---

$\diamondsuit$ $$\lim_{s \to -1} \frac{\psi(s)}{\Gamma(s) } = \lim_{s \to -1} \frac{-\frac{1}{s+1} + O(1)}{-\frac{1}{s+1}+O(1)} =1$$

---

Similarly, we have

$$\begin{align} \int_{0}^{\infty} \frac{e^{-x} \sin^{4}(x)}{x} \, \mathrm dx &= \frac{1}{8} \lim_{s \to 0} \Gamma(s) \left(3-\frac{4 \cos(s \arctan 2)}{5^{s/2}} + \frac{\cos(s \arctan 4)}{17^{s/2}}\right) \\ &= \frac{1}{8} \left( 2 \ln (5) - \frac{\ln (17)}{2} \right), \end{align} $$

$$ \begin{align} \int_{0}^{\infty} \frac{e^{-x} \sin^{4}(x)}{x^{3}} \, \mathrm dx &= \frac{1}{8} \lim_{s \to -2} \Gamma(s) \left(3-\frac{4 \cos(s \arctan 2)}{5^{s/2}} + \frac{\cos(s \arctan 4)}{17^{s/2}}\right) \\ &= \frac{1}{8} \left(- 8\arctan(2) - 3 \ln(5) + 4 \arctan(4) + \frac{15 \ln(17)}{4} \right), \end{align}$$

and $$ \begin{align} \int_{0}^{\infty} \frac{e^{-x} \sin^{4}(x)}{x^{4}} \, \mathrm dx &= \frac{1}{8} \lim_{s \to -3} \Gamma(s) \left(3-\frac{4 \cos(s \arctan 2)}{5^{s/2}} + \frac{\cos(s \arctan 4)}{17^{s/2}}\right) \\ &= \frac{1}{8} \left(- \frac{4\arctan(2)}{3} + \frac{11 \ln(5)}{3} + \frac{26 \arctan (4)}{3} - \frac{47 \ln(17)}{12}\right). \end{align}$$

Answer 4

There is no need to use Feynman trick.

$$I = \int_{0}^{\infty} \frac{e^{-x}\sin^4(x)}{x^{2}}\,dx$$

One integration by parts with $u=e^{-x}\sin^4(x)$ reduces the problem to $$I=\int_{0}^{\infty}\frac{e^{-x} \sin ^3(x) (4 \cos (x)-\sin (x))}{x}\,dx$$

Using Euler formulation of the sine and cosine function gives for the antiderivative $$J=\int\frac{e^{-x} \sin ^3(x) (4 \cos (x)-\sin (x))}{x}\,dx$$ $$J=\left(\frac{1+2i}{4}\right) \text{Ei}(-(1+2 i) x)+\left(\frac{1-2i}{4}\right) \text{Ei}(-(1-2 i) x)-$$ $$\left(\frac{1+4i}{16}\right) \text{Ei}(-(1+4 i) x)-\left(\frac{1-4i}{16}\right) \text{Ei}(-(1-4 i) x)-\frac{3 }{8}\text{Ei}(-x)$$

So, using the bounds, $$I=\frac \pi 2- \left(\frac{\pi }{4}+\frac{1}{16} \log \left(\frac{625}{17}\right)+\frac{1}{2} \tan ^{-1}\left(\frac{13}{16}\right)\right)$$ $$I=\frac{\pi }{4}-\frac{1}{16} \log \left(\frac{625}{17}\right)-\frac{1}{2} \tan ^{-1}\left(\frac{13}{16}\right)=0.218956$$

Answer 5

By one integration by parts $$I=\int_0^\infty \frac{e^{-x}\sin^4x}{x^2}dx=\int_0^\infty \frac{e^{-x}\sin^3x(4\cos x-\sin x)}xdx$$ and by using some trigonometric identities $$I=\int_0^\infty \frac{e^{-x}}x\left(\sin2x-\frac12\sin4x-\frac12(1-\cos2x)+\frac18(1-\cos8x)\right)dx. $$ Now we will use the integral identities below which can be proved by using an integral theorem that I don't remember (starts with F) $$\int_0^\infty \frac{e^{-x}\sin ax}xdx=\arctan a$$ $$\int_0^\infty\frac{e^{-x}(1-\cos ax)}x=\frac12\ln(a^2+1).$$ Hence $$I==\arctan2-\frac12\arctan4-\frac14\ln5+\frac1{16}\ln17. $$