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Question: Integration of \[\dfrac{{\sin x}}{x}\] from \[0\] to infinity?...

Integration of sin⁡xx\dfrac{{\sin x}}{x} from 00 to infinity?

Explanation

Solution

Hint : Here we will first consider the integration of a function which contains the given function. Then we will differentiate that function with respect to one of the variables, keeping the other as constant. Further we will integrate that derivative and equate it with the original function because the integral of the derivative gives the original function. Then we will find the value of constant and further we will find the integral ∫0∞sin⁡xxdx\int\limits_0^\infty {\dfrac{{\sin x}}{x}} dx.

Complete step-by-step answer :
Let,
I(s)=∫0∞sin⁡xxe−sxdxI\left( s \right) = \int\limits_0^\infty {\dfrac{{\sin x}}{x}} {e^{ - sx}}dx
Now we will differentiate it with respect to ss, keeping xx as constant.
⇒I′(s)=d[∫0∞sin⁡xxe−sxdx]ds\Rightarrow I'\left( s \right) = \dfrac{{d\left[ {\int\limits_0^\infty {\dfrac{{\sin x}}{x}{e^{ - sx}}dx} } \right]}}{{ds}}
We know that,
de−sxds=−xe−sx\dfrac{{d{e^{ - sx}}}}{{ds}} = - x{e^{ - sx}}
So, using this we will differentiate the integral. So, we get,
⇒I′(s)=∫0∞sin⁡xx(−x)e−sxdx\Rightarrow I'\left( s \right) = \int\limits_0^\infty {\dfrac{{\sin x}}{x}\left( { - x} \right){e^{ - sx}}dx}
Now we will cancel xx and take out the minus sign. So, we get,
⇒I′(s)=−∫0∞e−sxsin⁡xdx\Rightarrow I'\left( s \right) = - \int\limits_0^\infty {{e^{ - sx}}\sin x} dx
⇒−I′(s)=∫0∞e−sxsin⁡xdx\Rightarrow - I'\left( s \right) = \int\limits_0^\infty {{e^{ - sx}}\sin xdx}
Now we will use integration by parts to integrate the above integral.
First of all, we will number the functions using the ILATE rule. So, sin⁡x\sin x will be the first function and e−sx{e^{ - sx}} will be the second function. Now integrating according to the formula for integration by parts. We have;
⇒−I′(s)=sin⁡x∫e−sxdx−∫(dsin⁡xdx×∫e−sxdx)dx\Rightarrow - I'\left( s \right) = \sin x\int {{e^{ - sx}}dx - \int {\left( {\dfrac{{d\sin x}}{{dx}} \times \int {{e^{ - sx}}dx} } \right)} } dx
We know that,
∫e−sxdx=−e−sxs\int {{e^{ - sx}}dx = \dfrac{{ - {e^{ - sx}}}}{s}}
Using this in the above integral we get;
⇒−I′(s)=sin⁡x(−e−sxs)−∫cos⁡x×(−e−sxs)dx\Rightarrow - I'\left( s \right) = \sin x\left( {\dfrac{{ - {e^{ - sx}}}}{s}} \right) - \int {\cos x \times } \left( {\dfrac{{ - {e^{ - sx}}}}{s}} \right)dx
On rearranging we get;
⇒−I′(s)=−1se−sxsin⁡x+1s∫e−sxcos⁡xdx\Rightarrow - I'\left( s \right) = \dfrac{{ - 1}}{s}{e^{ - sx}}\sin x + \dfrac{1}{s}\int {{e^{ - sx}}\cos x} dx
Now for the second term on the RHS we will again carry out integration by parts in which the first function will be cos⁡x\cos x and the second function will be e−sx{e^{ - sx}}. So, we get;
⇒−I′(s)=−1se−sxsin⁡x+1s[cos⁡x∫e−sxdx−∫(dcos⁡xdx×∫e−sxdx)dx]\Rightarrow - I'\left( s \right) = \dfrac{{ - 1}}{s}{e^{ - sx}}\sin x + \dfrac{1}{s}\left[ {\cos x\int {{e^{ - sx}}dx - \int {\left( {\dfrac{{d\cos x}}{{dx}} \times \int {{e^{ - sx}}dx} } \right)dx} } } \right]
Again, we will use that;
∫e−sxdx=−e−sxs\int {{e^{ - sx}}dx = \dfrac{{ - {e^{ - sx}}}}{s}}
So, we have;
⇒−I′(s)=−1se−sxsin⁡x+1s[cos⁡x(−e−sxs)−∫−sin⁡x×(−e−sxs)dx]\Rightarrow - I'\left( s \right) = \dfrac{{ - 1}}{s}{e^{ - sx}}\sin x + \dfrac{1}{s}\left[ {\cos x\left( {\dfrac{{ - {e^{ - sx}}}}{s}} \right) - \int { - \sin x \times \left( {\dfrac{{ - {e^{ - sx}}}}{s}} \right)dx} } \right]
On further simplification we have;
⇒−I′(s)=−1se−sxsin⁡x+1s[cos⁡x(−e−sxs)−1s∫e−sxsin⁡xdx]\Rightarrow - I'\left( s \right) = \dfrac{{ - 1}}{s}{e^{ - sx}}\sin x + \dfrac{1}{s}\left[ {\cos x\left( {\dfrac{{ - {e^{ - sx}}}}{s}} \right) - \dfrac{1}{s}\int {{e^{ - sx}}\sin xdx} } \right]
Now we will take the terms out of the bracket. So, we have;
⇒−I′(s)=−1se−sxsin⁡x−1s2e−sxcos⁡x−1s2∫e−sxsin⁡xdx\Rightarrow - I'\left( s \right) = \dfrac{{ - 1}}{s}{e^{ - sx}}\sin x - \dfrac{1}{{{s^2}}}{e^{ - sx}}\cos x - \dfrac{1}{{{s^2}}}\int {{e^{ - sx}}\sin xdx}
Now we will put the limits, so we have;
⇒−I′(s)=[−1se−sxsin⁡x−1s2e−sxcos⁡x]0∞−1s2∫0∞e−sxsin⁡xdx\Rightarrow - I'\left( s \right) = \left[ {\dfrac{{ - 1}}{s}{e^{ - sx}}\sin x - \dfrac{1}{{{s^2}}}{e^{ - sx}}\cos x} \right]_0^\infty - \dfrac{1}{{{s^2}}}\int\limits_0^\infty {{e^{ - sx}}\sin xdx}
Now we can see that the integral in the RHS is the same integral that we have assumed to be equal to I′(s)I'\left( s \right).
So, we can write;
⇒−I′(s)=[−1se−sxsin⁡x−1s2e−sxcos⁡x]0∞+I′(s)s2\Rightarrow - I'\left( s \right) = \left[ {\dfrac{{ - 1}}{s}{e^{ - sx}}\sin x - \dfrac{1}{{{s^2}}}{e^{ - sx}}\cos x} \right]_0^\infty + \dfrac{{I'\left( s \right)}}{{{s^2}}}
Now we will shift the last term to LHS.
⇒(−1−1s2)I′(s)=[−1se−sxsin⁡x−1s2e−sxcos⁡x]0∞\Rightarrow \left( { - 1 - \dfrac{1}{{{s^2}}}} \right)I'\left( s \right) = \left[ {\dfrac{{ - 1}}{s}{e^{ - sx}}\sin x - \dfrac{1}{{{s^2}}}{e^{ - sx}}\cos x} \right]_0^\infty
Now we will put the limits.
⇒(−s2−1s2)I′(s)=[−1se−∞ssin⁡∞−1s2e−∞scos⁡∞]−[−1se−0ssin⁡0−1s2e−0scos⁡0]\Rightarrow \left( {\dfrac{{ - {s^2} - 1}}{{{s^2}}}} \right)I'\left( s \right) = \left[ {\dfrac{{ - 1}}{s}{e^{ - \infty s}}\sin \infty - \dfrac{1}{{{s^2}}}{e^{ - \infty s}}\cos \infty } \right] - \left[ {\dfrac{{ - 1}}{s}{e^{ - 0s}}\sin 0 - \dfrac{1}{{{s^2}}}{e^{ - 0s}}\cos 0} \right]
We know that;
e−∞s=0{e^{ - \infty s}} = 0
sin⁡∞\sin \infty is a number oscillating between −1,1 - 1,1.
cos⁡∞\cos \infty is also a number oscillating between −1,1 - 1,1.
So,
\Rightarrow $$$${e^{ - \infty s}}\cos \infty = 0
\Rightarrow $$$${e^{ - \infty s}}\sin \infty = 0.
So, we get,
⇒(−s2−1s2)I′(s)=[0−0]−[0−1s2]\Rightarrow \left( {\dfrac{{ - {s^2} - 1}}{{{s^2}}}} \right)I'\left( s \right) = \left[ {0 - 0} \right] - \left[ {0 - \dfrac{1}{{{s^2}}}} \right]
⇒(−s2−1s2)I′(s)=1s2\Rightarrow \left( {\dfrac{{ - {s^2} - 1}}{{{s^2}}}} \right)I'\left( s \right) = \dfrac{1}{{{s^2}}}
So, on cancelling the terms and shifting we get;
⇒I′(s)=1−s2−1\Rightarrow I'\left( s \right) = \dfrac{1}{{ - {s^2} - 1}}
Taking the minus sign common we get;
⇒I′(s)=−1s2+1\Rightarrow I'\left( s \right) = \dfrac{{ - 1}}{{{s^2} + 1}}
Now we will integrate this with respect to ss. So, we get;
⇒∫I′(s)ds=∫−1s2+1ds\Rightarrow \int {I'\left( s \right)ds} = \int {\dfrac{{ - 1}}{{{s^2} + 1}}} ds
Now we know that; ∫1x2+1dx=tan⁡−1x+c\int {\dfrac{1}{{{x^2} + 1}}} dx = {\tan ^{ - 1}}x + c
So, using this we get,
⇒I(s)=−tan⁡−1s+c\Rightarrow I\left( s \right) = - {\tan ^{ - 1}}s + c
But we have;
I(s)=∫0∞sin⁡xxe−sxdxI\left( s \right) = \int\limits_0^\infty {\dfrac{{\sin x}}{x}} {e^{ - sx}}dx
So, both of them are equal and hence we have,
⇒−tan⁡−1s+c=∫0∞sin⁡xxe−sxdx\Rightarrow - {\tan ^{ - 1}}s + c = \int\limits_0^\infty {\dfrac{{\sin x}}{x}} {e^{ - sx}}dx
Now to find the value of constant we will put s=∞s = \infty . So, we have
⇒−tan⁡−1∞+c=∫0∞sin⁡xxe−∞xdx\Rightarrow - {\tan ^{ - 1}}\infty + c = \int\limits_0^\infty {\dfrac{{\sin x}}{x}} {e^{ - \infty x}}dx
Now we know, e−∞s=0{e^{ - \infty s}} = 0 and tan⁡−1∞=π2{\tan ^{ - 1}}\infty = \dfrac{\pi }{2}.
So, the whole term on the RHS becomes zero.
⇒−π2+c=0\Rightarrow - \dfrac{\pi }{2} + c = 0
⇒c=π2\Rightarrow c = \dfrac{\pi }{2}
Now we will put the value of cc. So, we get;
⇒∫0∞sin⁡xxe−sxdx=−tan⁡−1s+π2\Rightarrow \int\limits_0^\infty {\dfrac{{\sin x}}{x}} {e^{ - sx}}dx = - {\tan ^{ - 1}}s + \dfrac{\pi }{2}
Now we will put s=0s = 0.
⇒∫0∞sin⁡xxe−0xdx=−tan⁡−10+π2\Rightarrow \int\limits_0^\infty {\dfrac{{\sin x}}{x}} {e^{ - 0x}}dx = - {\tan ^{ - 1}}0 + \dfrac{\pi }{2}
Now we know, e0=1{e^0} = 1 and tan⁡−10=0{\tan ^{ - 1}}0 = 0.
So, we get,
⇒∫0∞sin⁡xxdx=π2\Rightarrow \int\limits_0^\infty {\dfrac{{\sin x}}{x}} dx = \dfrac{\pi }{2}
So, the correct answer is “π2 \dfrac{\pi }{2}”.

Note : One important thing to note here is that while performing integration by parts method we should always choose the first or second function by the ILATE rule. Here, I stand for inverse trigonometric function, L stands for logarithmic function, A stands for algebraic function, T stands for trigonometric function, E stands for exponential function. We can also solve this question by the use of Fourier transform.