Solve Strange Integral of (sinx)^2/x^2 from -inf to inf

du=2sin(x)cos(x)dv=1/x^2sointegral of (sinx)^2 / (x^2) from -infinity to infinity = uv - integral of v du=-sin(x)^2/x - integral of (-2sin(x)cos(x)/x)(-1/x) dx=-sin(x)^2/x + 2integral of sin(x)/x dx=-sin(x)^2/x + 2integral of sin(x)/x dx=2pi
  • #1
kryoo
2
0
integral of (sinx)^2 / (x^2) from -infinity to infinity. how to tackle this
 
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  • #2
I looked it up in a table. The answer is π.
 
  • #3
If you want to work it out you could use differentiation/integration with respect to a parametor, laplace transforms, complex integration and the residue theorem, or it can be reduced to the integral of sin(x)/x by integration by parts.
 
  • #4
lurflurf said:
If you want to work it out you could use differentiation/integration with respect to a parametor, laplace transforms, complex integration and the residue theorem, or it can be reduced to the integral of sin(x)/x by integration by parts.

so it's pi?

how do you reduce it to sinx/x with integration by parts, using resides I got 2pi
 
  • #5
it is pi

u dv=u v-v du
let
u=sin(x)^2
v=-1/x
 

1. What is the value of the integral?

The value of the integral is 0.

2. How do you solve this integral?

To solve this integral, we can use the substitution method. Let u = x and du = dx. This will transform the integral to ∫(sinu)^2/u^2 from -inf to inf. Then, we can use the trigonometric identity (sinx)^2 = (1-cos2x)/2 to simplify the integral to ∫(1-cos2u)/2u^2. This can be further simplified to ∫(1/2u^2)-cos2u/2u^2. Using the power rule for integration and the integral of cosine, we get the final result of 1/2(pi/2) - 0, which is equal to 0.

3. Can this integral be solved without using substitution?

No, this integral cannot be solved without using substitution. The integral involves a trigonometric function with a variable in the denominator, which cannot be solved using basic integration rules. Substitution is necessary to simplify the integral and solve it.

4. What is the significance of the limits of integration being from -inf to inf?

The limits of integration being from -inf to inf indicate that the integral is over the entire real number line. This means that the integral takes into account both positive and negative values of x, which is important when dealing with trigonometric functions.

5. Is this integral used in any real-world applications?

Yes, this integral is used in various areas of physics and engineering, particularly in the study of oscillating systems. The integral helps in calculating the energy of a vibrating system and determining its stability.

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