I'm trying to find(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\int\limits_0^{ + \infty } {\frac{{\sin x}}

{{x^3 + x}}dx} [/tex]

Since the function is even, I can compute it as

[tex]\frac{1}

{2}\int\limits_{ - \infty }^{ + \infty } {\frac{{\sin x}}

{{x^3 + x}}dx} [/tex]

To use the residue theorem, I construct a large semi-circle C with center O and radius R. To avoid the real singularity at x = 0, we construct another (small) semi-circle c with radius r. The closed path is then given by C + [-R,-r] + c + [r,R]. We then let R go to infinity and r to 0. So we have:

[tex]\oint {\frac{{\sin x}}

{{x^3 + x}}dx} = \int\limits_C {\frac{{\sin x}}

{{x^3 + x}}dx} + \int\limits_c {\frac{{\sin x}}

{{x^3 + x}}dx} + \int\limits_{ - R}^{ - r} {\frac{{\sin x}}

{{x^3 + x}}} dx + \int\limits_r^R {\frac{{\sin x}}

{{x^3 + x}}} dx[/tex]

From the residue theorem, I found the LHS to be [itex]\pi \sinh 1[/itex], by applying the theorem to the only enclosed pole i.

Then, taking the limits or R and r to inf and 0, I found the integral over C to be 0 and over c to be [itex]-\pi[/itex] which results in:

[tex]\int\limits_{ - \infty }^{ + \infty } {\frac{{\sin x}}

{{x^3 + x}}} dx = \pi \sinh 1 + \pi [/tex]

Now that appears to be too large, does anyone know what mistake(s) I made or what I should do? Thanks!

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Improper integral using the Residue Theorem

**Physics Forums | Science Articles, Homework Help, Discussion**