# Thread: Integrals #6 and #7

1. ## Integrals #6 and #7

Challenge Problems:

$\displaystyle \int^{b}_{a} \frac{e^{x /a} - e^{b /x}}{x} \ dx$

$\displaystyle \int^{1}_{0} \sqrt[3]{2x^{3}-3x^2-x+1} \ dx$

EDIT: I don't like the second integral. It doesn't even make any sense unless you state that $\displaystyle \sqrt[3]{2x^{3}-3x^2-x+1}$ is a real-valued function. Sorry. It's a bad problem.

Moderator edit: Approved Challenge question.

2. Originally Posted by Random Variable
Challenge Problems:

$\displaystyle \int^{b}_{a} \frac{e^{x /a} - e^{b /x}}{x} \ dx$
$\displaystyle \int^{b}_{a} \frac{e^{x /a} - e^{b /x}}{x} \ dx =0$

A simple Taylor/Laurent expansion of $\displaystyle e^{x /a}$ and $\displaystyle e^{b /x}$ should prove my assertion.

3. Originally Posted by chiph588@
$\displaystyle \int^{b}_{a} \frac{e^{x /a} - e^{b /x}}{x} \ dx =0$

A simple Taylor/Laurent expansion of $\displaystyle e^{x /a}$ and $\displaystyle e^{b /x}$ should prove my assertion.
Could you elaborate?

The solution I had in mind is even simpler than that.

4. If my calculations are correct , the integrals are both zero .

For the first one ,

Sub. $\displaystyle x = \frac{ab}{t}$ in the second integral .

and we will find that it is equal to

$\displaystyle \int_a^b \frac{e^{x/a}}{x}~dx - \int_a^b \frac{e^{t/a}}{t}~dt$

And the second integral , i find that it is identical to

$\displaystyle \sqrt[3]{(2x-1)[(x-\frac{1}{2})^2 - \frac{5}{4}]}$

Sub. $\displaystyle x = 1-t$ and we have

$\displaystyle I = -I$

$\displaystyle I = 0$

5. Originally Posted by simplependulum
If my calculations are correct , the integrals are both zero .

For the first one ,

Sub. $\displaystyle x = \frac{ab}{t}$ in the second integral .

and we will find that it is equal to

$\displaystyle \int_a^b \frac{e^{x/a}}{x}~dx - \int_a^b \frac{e^{t/a}}{t}~dt$

And the second integral , i find that it is identical to

$\displaystyle \sqrt[3]{(2x-1)[(x-\frac{1}{2})^2 - \frac{5}{4}]}$

Sub. $\displaystyle x = 1-t$ and we have

$\displaystyle I = -I$

$\displaystyle I = 0$

Nice.

If for the first integral you made the same substitution but didn't break it up into two integrals, you would get I = -I.

And for the second integral, Maple claims the answer is a complex number and Mathematica doesn't know what to do with it. The issue is the cube root of negative real numbers.

6. Originally Posted by Random Variable
$\displaystyle \int^{1}_{0} \sqrt[3]{2x^{3}-3x^2-x+1} \ dx$
Let $\displaystyle x=t+\tfrac12$

We get $\displaystyle \int^{1}_{0} \sqrt[3]{2x^{3}-3x^2-x+1} \ dx = \int^{\tfrac12}_{-\tfrac12} \sqrt[3]{2(t+\tfrac12)^{3}-3(t+\tfrac12)^2-(t+\tfrac12)+1} \ dx =$ $\displaystyle \int^{\tfrac12}_{-\tfrac12} \sqrt[3]{2t^3-\tfrac52t} \ dx = 0$ since $\displaystyle \sqrt[3]{2t^3-\tfrac52t}$ is odd.