Evaluating the integral $\int_{-1}^{1} \frac{\sin{x}}{1+x^2}dx$

Question

I was asked to evaluate the integral

$$\int_{-1}^{1} \frac{\sin{x}}{1+x^2}dx$$

if it exists.

This is a problem from Calculus and the student has been taught how to use trigonometric substitution. My intuition was to do trig sub with $$x=\tan{\theta}$$ and eliminating $$\frac{dx}{1+x^2}$$ but when I do that the numerator becomes $$\sin{(\tan{\theta})}$$ which I am not comfortable integrating.

Another queue was that the problem asks "if it exists" so I tried to see if there are any points within $(-1,1)$ that may cause any problem, but I don't think I see any asymptotes or undefined numbers so I'm not sure if I am dealing with an improper integral.

Can someone help me out on this? Thank you.

Jeff Faraci · Accepted Answer · 2014-05-14T01:27:00.040

12

Zero by symmetry.

It is an odd integrand integrated over a symmetric bound

or more properly "integrated over an interval that is symmetric about the origin".

edited May 14 '14 at 01:27

answered May 14 '14 at 01:19

Jeff Faraci

10,393

1

Maybe say integrated on an interval symmetric about the origin? – Ted Shifrin May 14 '14 at 01:23
@TedShifrin Thank you it is not my first language english, you helped me. I added it. – Jeff Faraci May 14 '14 at 01:27
No problem. Even for native English speakers, math is often tricky :) – Ted Shifrin May 14 '14 at 01:31
It sure is! Thank you Ted – Jeff Faraci May 14 '14 at 01:32
@Integrals What would you say $\int_{-\infty}^\infty xdx$ is? – Gamma Function May 14 '14 at 02:42
@Integrals My point: You must also check for convergence. – Gamma Function May 14 '14 at 02:44
2

Do you know what differential equation you belong to? See my post earlier today that deals with convergence in full detail for these kinds of integrals . Thank you. http://math.stackexchange.com/questions/793595/convergence-i-int-0-infty-frac-sin-xxsdx/793609#793609 @GammaFunction – Jeff Faraci May 14 '14 at 02:44
I am searching for it too if I find it I shall tell you @GammaFunction – Jeff Faraci May 14 '14 at 02:47
@Integrals There you go, what a lovely solution x2. – Gamma Function May 14 '14 at 02:50
3

Wow, it's so funny how blind a person can be when there is such an obvious answer right in front of you. Thanks. – hyg17 May 14 '14 at 03:12
@hyg17 No problem...Thank you we are all blind in math – Jeff Faraci May 14 '14 at 17:24

score 5 · Answer 2 · answered May 14 '14 at 01:28

The integral is \begin{align} I &= \int_{-1}^{1} \frac{\sin(x)}{1+x^{2}} dx \\ &= \int_{-1}^{0} \frac{\sin(x)}{1+x^{2}} dx + \int_{0}^{1} \frac{\sin(x)}{1+x^{2}} dx \\ &= - \int_{0}^{1} \frac{\sin(x)}{1+x^{2}} dx + \int_{0}^{1} \frac{\sin(x)}{1+x^{2}} dx \\ &= 0 \end{align} where the change of variable $x \rightarrow -x$ was made.

score 2 · Answer 3 · answered May 14 '14 at 01:45

We will begin by evaluating $$I(x) = \int\!\dfrac{\sin(x)}{1+x^2}\mathrm{d}x$$

$$I = \dfrac{1}{2i} \int\!\dfrac{\sin(x)}{x-i}\mathrm{d}x - \dfrac{1}{2i}\int\!\dfrac{\sin(x)}{x+i}\mathrm{d}x$$

We will now consider $$J_\pm = \int\!\dfrac{\sin(x)}{x\pm i}\mathrm{d}x$$

Substitute $u = x \pm i$.

$$J_\pm = \int\!\dfrac{\sin(u \mp i)}{u}\mathrm{d}x = \int\!\dfrac{\sin(u)\cos(i)\mp\sin(i)\cos(u)}{u}\mathrm{d}x$$

Splitting the integral and using hyperbolic functions

$$J_\pm = \int\!\dfrac{\sin(u)\cosh(1)}{u}\mathrm{d}x \mp\int\!i\dfrac{\sinh(1)\cos(u)}{u}\mathrm{d}x$$

Writing in terms of $\operatorname{Si}(x)$ and $\operatorname{Ci}(x)$

$$J_\pm = \cosh(1)\operatorname{Si}(u) \mp i\sinh(1)\operatorname{Ci}(u)$$

$$J_\pm = \cosh(1)\operatorname{Si}(x\pm i) \mp i\sinh(1)\operatorname{Ci}(x\pm i)$$

$$I = \dfrac{1}{2i}\left(J_--J_+\right)$$

$$I = \dfrac{\left(\cosh(1)\operatorname{Si}(x- i) + i\sinh(1)\operatorname{Ci}(x- i)\right)-\left(\cosh(1)\operatorname{Si}(x+ i) - i\sinh(1)\operatorname{Ci}(x+ i)\right)}{2i}$$

Now evaluate $I$ at $1$ and $-1$.

$$I(1) = I(-1) = -0.324967578038053553$$

$$I(1) - I(-1) = \boxed{0}$$

If you are sneaky you can use symmetry to show that $I(1) = I(-1)$ without actually calculating the value.

@TedShifrin but it is useful to everyone else. The problem is posted for the benefit of the community rather than simply the person who asks the question. This is why "defacing" your own questions is not allowed. If anyone wants to see how they would evaluate this indefinite integral without W|A they can refer to my post instead of posting a new thread. An "acceptable" answer has already been posted. — Brad, May 14 '14 at 01:50

Evaluating the integral $\int_{-1}^{1} \frac{\sin{x}}{1+x^2}dx$

3 Answers3