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Problem: Integration of $\displaystyle\int_{-1}^1 {\sin x\over 1+x^2} \; dx = 0 $

(according to WolframAlpha Definite Integral Calculator)

But I don't understand how. I tried to prove using integration by parts. Here's the work: $$ \int_{-1}^1 {\sin x\over 1+x^2} \; dx = \int_{-1}^1 {\sin x}{1\over 1+x^2} \; dx $$

Let $u = \sin x,\quad du = \cos x\; dx\;$ and $v = \tan^{-1}x,\quad dv = {1\over 1+x^2}dx\;$. So $$ \int_{-1}^1 u dv = \left[uv\right]_{-1}^1 - \int_{-1}^1 v du =\left[ \sin x (\tan^{-1}x)\right]_{-1}^1 - \int_{-1}^1 \tan^{-1}x \cos x\; dx. $$

Next let $u = \tan^{-1}x, du = {1\over 1+x^2}$ and $dv = \cos x, v = \sin x$...

I stopped here, because I feel like I'm going in a circle with this problem. What direction would I take to solve this because I don't know whether integration by parts is the way to go? Should I use trig substitution?


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The fact that we have an odd function does it. If you really insist on (sort of) integrating, look at the integral from $-1$ to $0$, make the change of variable $u=-x$. After a short time you should get $\int_{-1}^0 \frac{\sin x}{1+x^2}\,dx=-\int_{0}^1 \frac{\sin u}{1+u^2}\,du$, which, since the variable name doesn't matter, is just $-\int_0^1\frac{\sin x}{1+x^2}\,dx$. – André Nicolas Apr 2 '12 at 4:53
That second integration by parts would reverse the first one. I have no idea how you'd do the indefinite integral for this one. I don't even think the Weierstrass substitution would save you on this one. – Mike Apr 2 '12 at 6:11
The indefinite integral is a non-elementary function which can be expressed in terms of the special functions Si and Ci. The standard techniques of first-year calculus will get you nowhere. – Robert Israel Apr 2 '12 at 7:29
I edited your question. Please check, if everthings still as you wanted it! – draks ... Apr 2 '12 at 8:15
@draks that is what I should have written. thanks. (I'm in Calculus II) and I do not feel like my level helps me answer this. Thank you for all the responses, it gives me some insight into how to solve it though. – fragilewindows Apr 2 '12 at 13:59

You don’t have to do any actual integration. Let $$f(x)=\frac{\sin x}{1+x^2}\;$$ then $$f(-x)=\frac{\sin(-x)}{1+(-x)^2}=\frac{-\sin x}{1+x^2}=-f(x)\;,$$ so $f(x)$ is an odd function. The signed area between $x=-1$ and $x=0$ is therefore just the negative of the signed area from $x=0$ to $x=1$, and the whole thing cancels out.

In more detail, let $$A=\int_0^1 f(x) dx=\int_0^1\frac{\sin x}{1+x^2} dx\;,$$ and let $$B=\int_{-1}^0 f(x) dx=\int_{-1}^0\frac{\sin x}{1+x^2} dx\;.\tag{1}$$ Now substitute $u=-x$ in $(1)$: $f(u)=f(-x)=-f(x)$, $du=-dx=(-1)dx$ so $dx=-du$, and $u$ runs from $1$ to $0$, so

$$B=\int_1^0 -f(x)(-1)dx=\int_1^0f(x)dx=-\int_0^1f(x)dx=-A\;.$$

Thus, $$\int_{-1}^1 f(x)dx=A+B=A-A=0\;.$$

Note that the specific function $f$ didn’t matter: we used only the fact that $f$ is an odd function.

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so are we comparing the equation at x = 1 and x = -1? Also, how do we know f(x) is an odd function? (sorry if this is a silly question) I understand 90% of the explanations provided – fragilewindows Apr 2 '12 at 6:18
@fragilewindows: The second sentence of my answer is a proof that $f$ is odd; remember, that simply means that $f(-x)=-f(x)$ for all $x$. No, we’re not just comparing $f(1)$ and $f(-1)$: we’re using the fact that the graph of the function from $-1$ to $0$ is the inverted mirror image of the graph from $0$ to $1$. If you have a graphing calculator, use it to display $y=f(x)$ from $-1$ to $1$, and you should see what I mean: the left side dips below the axis as you go from $0$ to $-1$ in exactly the same way that the right side rises above it as you go from $0$ to $1$. – Brian M. Scott Apr 2 '12 at 6:23

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