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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?

Thanks.

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    @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.2012-04-02

1 Answers 1

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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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    @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 $1$.2012-04-02