# Some impossible integrals...

• Nov 16th 2012, 05:15 AM
TibiSam
Some impossible integrals...
Hy everyone!
I don't know exactly how to use some symbols here, but i will denote with int(function) as being the integral of a function.
So I have 4 problems, each of them will be very helpful if you can give a suggestion somehow...

1) lim (n tends to infinity) of n^2 * ∫∫((x^2-y^2)/4)^n dx dy.
The first integral is defined from 0 to 1 (and the variables is y, and the second from y to 2-y and the variable is x).
There is also a hint given: change variables by rotating the axes through an angle of pi/4.

2) ∫∫e^(2x-x^2) dx dy. The first integral is defined from 0 to 1 and the variable is y, the second is defined from 0 to 1-y(1/3) and the variable is x.

3) ∫ 1/((1+x^2)*(1+x^a)) dx, where a is a constant. The integral is defined from 0 to infinity, and the variable is x.

4) ∫f * (x-1/x) dx, where f is a probability density function, the integral is defined from minus infinity to plus infinity and the variable is x.

Thank you a lot if you manage to help me somehow.
• Nov 16th 2012, 07:22 AM
Prove It
Re: Some impossible integrals...
Quote:

Originally Posted by TibiSam
Hy everyone!
I don't know exactly how to use some symbols here, but i will denote with int(function) as being the integral of a function.
So I have 4 problems, each of them will be very helpful if you can give a suggestion somehow...

1) lim (n tends to infinity) of n^2 * ∫∫((x^2-y^2)/4)^n dx dy.
The first integral is defined from 0 to 1 (and the variables is y, and the second from y to 2-y and the variable is x).
There is also a hint given: change variables by rotating the axes through an angle of pi/4.

2) ∫∫e^(2x-x^2) dx dy. The first integral is defined from 0 to 1 and the variable is y, the second is defined from 0 to 1-y(1/3) and the variable is x.

3) ∫ 1/((1+x^2)*(1+x^a)) dx, where a is a constant. The integral is defined from 0 to infinity, and the variable is x.

4) ∫f * (x-1/x) dx, where f is a probability density function, the integral is defined from minus infinity to plus infinity and the variable is x.

Thank you a lot if you manage to help me somehow.

For number 2, I would suggest reversing the order of integration, as \displaystyle \begin{align*} \int{e^{2x - x^2}\,dx} \end{align*} can not be evaluated.

Your integration limits give \displaystyle \begin{align*} 0 \leq y \leq 1 \end{align*} and \displaystyle \begin{align*} 0 \leq x \leq 1 - y^{\frac{1}{3}} \end{align*}. Some rearranging gives \displaystyle \begin{align*} 0 \leq x \leq 1 \end{align*} and \displaystyle \begin{align*} 0 \leq y \leq (1 - x)^3 \end{align*}, so your integral is

\displaystyle \begin{align*} \int_{0}^{1}{\int_{0}^{1 - y^{\frac{1}{3}}}{e^{2x - x^2}\,dx}\,dy} &= \int_{0}^{1}{\int_{0}^{(1 - x)^3}{e^{2x - x^2}\,dy}\,dx} \\ &= \int_0^1{\left[ y\,e^{2x - x^2} \right]_{0}^{(1 - x)^3}\,dx} \\ &= \int_0^1{(1 - x)^3\, e^{2x - x^2}\,dx} \\ &= \int_0^1{(1 - x)^2(1 - x)\, e^{2x - x^2}\,dx} \\ &= \frac{1}{2}\int_0^1{(1 - x)^2(2 - 2x)\,e^{2x - x^2}\,dx} \end{align*}

Now make the substitution \displaystyle \begin{align*} u = e^{2x - x^2} \implies du = (2 - 2x)\,e^{2x - x^2} \end{align*}, note that when \displaystyle \begin{align*} x = 0, u = 1 \end{align*} and when \displaystyle \begin{align*} x = 1, u = e \end{align*} and also that
\displaystyle \begin{align*} u &= e^{2x - x^2} \\ \ln{(u)} &= 2x - x^2 \\ -\ln{(u)} &= x^2 - 2x \\ 1 - \ln{(u)} &= x^2 - 2x + 1 \\ 1 - \ln{(u)} &= (x - 1)^2 \end{align*}
then the integral becomes

\displaystyle \begin{align*} \frac{1}{2}\int_0^1{(1 - x)^2(2 - 2x)\,e^{2x - x^2}\,dx} &= \frac{1}{2}\int_1^e{1 - \ln{(u)}\,du} \\ &= \frac{1}{2}\left\{ u - \left[ u\ln{(u)} - u \right] \right\}_1^e \\ &= \frac{1}{2}\left[ 2u - u\ln{(u)} \right]_1^e \\ &= \frac{1}{2} \left\{ \left[ 2e - e\ln{(e)} \right] - \left[ 2\cdot 1 - 1 \ln{(1)} \right] \right\} \\ &= \frac{1}{2} \left( e - 2 \right) \end{align*}
• Nov 16th 2012, 08:22 AM
TibiSam
Re: Some impossible integrals...
Thanks a lot!