5
$\begingroup$

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.

$\endgroup$
12
$\begingroup$

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".

$\endgroup$
11
  • 1
    $\begingroup$ Maybe say integrated on an interval symmetric about the origin? $\endgroup$ – Ted Shifrin May 14 '14 at 1:23
  • $\begingroup$ @TedShifrin Thank you it is not my first language english, you helped me. I added it. $\endgroup$ – Jeff Faraci May 14 '14 at 1:27
  • $\begingroup$ No problem. Even for native English speakers, math is often tricky :) $\endgroup$ – Ted Shifrin May 14 '14 at 1:31
  • 2
    $\begingroup$ 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. math.stackexchange.com/questions/793595/… @GammaFunction $\endgroup$ – Jeff Faraci May 14 '14 at 2:44
  • 3
    $\begingroup$ Wow, it's so funny how blind a person can be when there is such an obvious answer right in front of you. Thanks. $\endgroup$ – hyg17 May 14 '14 at 3:12
5
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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

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

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

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.

$\endgroup$
3
  • $\begingroup$ This is not of use to a beginning calculus student. $\endgroup$ – Ted Shifrin May 14 '14 at 1:48
  • $\begingroup$ @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. $\endgroup$ – Brad May 14 '14 at 1:50
  • $\begingroup$ This was actually very helpful for me. Thanks. $\endgroup$ – hyg17 May 14 '14 at 3:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.