Determing if a function with integral within it is even/odd

[mr.tea]

Homework Statement

Determine if the function is even/odd/neither without solving the integral

Homework Equations

f(x)=\arctan (x)-2\int_0 ^x{\frac{1}{(1+t^2)^2}}

The Attempt at a Solution

I tried to do f(-x)-f(x). I know that if f is odd, I should get -2f(x) and if even, 0. I got that f is odd(-2f(x)), but I don't know if this is a correct answer, since wolfram alpha says that this is not an odd function. I didn't assume anything on the function while checking this. I am a bit confused.
Is there any way to check things like that?

Thank you,
Thomas

 

[Khashishi]

When the limits of the integration are 0 to x, the integral of an even function is odd and the integral of an odd function is even.
Look at each term in the equation. If each part is even, then the whole thing is even. Same with odd. But if you add an even and odd function (that are not 0) then you get neither.

 

[HallsofIvy]

First you know that arctan(x) is an odd function, right? Further, making the change of variable, u= -t in the integral, so that dt= -du
\int_0^x \frac{dt}{(1+ t^2)^2}
becomes
\int_0^{-x}\frac{-du}{(1+ (-u)^2)^2}= -\int_0^{-x} \frac{du}{(1+ u^2)^2}
changing the "dummy variable", u, in that integral to t,
\int_0^x \frac{dt}{(1+ t^2)^2}= -\int_0^{-x} \frac{dt}{(1+ t^2)^2}

 

[mr.tea]

When the limits of the integration are 0 to x, the integral of an even function is odd and the integral of an odd function is even.
Look at each term in the equation. If each part is even, then the whole thing is even. Same with odd. But if you add an even and odd function (that are not 0) then you get neither.

Hi,
First, thank you for the answer.
How do we know that the integral of an odd function is even and the same with even function? Maybe I will try to prove it, but expect to a question about that :)

Second, since the integrand is even, that means(according to you, but I am still not convinced about it) that the integral is odd. And since arctan is odd, that means that the whole function is odd. Right?

 

[epenguin]

Hi,
First, thank you for the answer.
How do we know that the integral of an odd function is even and the same with even function? Maybe I will try to prove it, but expect to a question about that :)

Second, since the integrand is even, that means(according to you, but I am still not convinced about it) that the integral is odd. And since arctan is odd, that means that the whole function is odd. Right?

HoI essentially just answered your question in #4. He did it for the particular function, but you could write out the proof again for integral of any function f which satisfies the condition of evenness which you should write out. The key point which you may have missed in the detail is, and may even disbelieve till you think about it is, to say it crudely

∫_a^b = - ∫_b^a

 

[mr.tea]

First you know that arctan(x) is an odd function, right? Further, making the change of variable, u= -t in the integral, so that dt= -du
\int_0^x \frac{dt}{(1+ t^2)^2}
becomes
\int_0^{-x}\frac{-du}{(1+ (-u)^2)^2}= -\int_0^{-x} \frac{du}{(1+ u^2)^2}
changing the "dummy variable", u, in that integral to t,
\int_0^x \frac{dt}{(1+ t^2)^2}= -\int_0^{-x} \frac{dt}{(1+ t^2)^2}

Thank you. I did something similar with the function but without changing the variable.

HoI essentially just answered your question in #4. He did it for the particular function, but you could write out the proof again for integral of any function f which satisfies the condition of evenness which you should write out. The key point which you may have missed in the detail is, and may even disbelieve till you think about it is, to say it crudely

∫_a^b = - ∫_b^a

As I said to HallsofIvy, I did the same thing. I just assumed that without loss of generality, x is positive then -x is negative and you flip the limits of the integral and you add minus.

I cannot see those things/"proofs" in my head without write them. That might be the reason I was( well, and still a bit) sceptical(Well, I am studying mathematics, I guess I should be sceptical).

So, I understand that my answer is correct :)

Thank you all.