Impossible question? (1 Viewer)

leehuan · May 19, 2016

Paradoxica said:
Quite so, according to Wolfram Alpha.

...Oh wow

So is the answer then
inf -> x>0
NaN -> x=0
-inf -> x<0

SammyT123 · May 19, 2016

InteGrand · May 19, 2016

SammyT123 said:
$When evaluating$ \int^{pi/4}_{0} \frac{ln2}{2} \\ \\ $How do we evaluate ln0?$

The ln(2)/2 is just a constant, so integrating that gives (pi/4)*(ln(2)/2) = pi*(ln(2)/8).

$\noindent In general, if $c$ is a constant, then $\int _a ^b c \text{ d}x = c\left(b-a \right).$

SammyT123 · May 19, 2016

Also from the same booklet

$\int^{e}_{1/e}\frac{\arctan(x)dx}{x}$

I just tried IBP, and used the substitution $u=ln(1/x) $ for J$$

This gave me

$I = (\tan^{-1}{x}ln{x})^{e}_{1/e} - J \\ J= \int^{1}_{0}\frac{-udu}{e^u+e^{-u}}$

SammyT123 · May 19, 2016

InteGrand said:
The ln(2)/2 is just a constant, so integrating that gives (pi/4)*(ln(2)/2) = pi*(ln(2)/8).

$\noindent In general, if $c$ is a constant, then $\int _a ^b c \text{ d}x = c\left(b-a \right).$

Whoops, messy writing on my part had my 'x' inside the ln()

InteGrand · May 19, 2016

SammyT123 said:
Also from the same booklet

$\int^{e}_{1/e}\frac{\arctan(x)dx}{x}$

I just tried IBP, and used the substitution $u=ln(1/x) $ for J$$

This gave me

$I = (\tan^{-1}{x}ln{x})^{e}_{1/e} - J \\ J= \int^{1}_{0}\frac{-udu}{e^u+e^{-u}}$

$\noindent This is another one that will be based on symmetry. Here's a sketch of the method. Write $I(a) = \int _{\frac{1}{a}}^{a} \frac{\tan^{-1} x}{x} \text{ d}x$, for $a\in \mathbb{R}$. (Note $a\neq 0$.) First deal with the case where $a>0$. Use a substitution of $x=\frac{1}{u}$ and use the identity $\tan^{-1} \frac{1}{\xi}= \frac{\pi}{2}-\tan^{-1} \xi$ for $\xi >0$. You should eventually end up with an equation involving $I(a)$, (because after doing the substitution, $I(a)$ will end up popping out again) which you can solve to find $I(a)$ (so similarly to before, we won't end up finding a primitive to the given integrand, which doesn't have an elementary primitive anyway. The reason we can do this integral is due to the nature of the bounds and the inverse tan identity).$

$\noindent Once we've managed to find an expression for $I(a)$ for $a>0$, we can find one for negative $a$ using symmetry of the bounds and the integrand, which is even. Note that for negative $a$, the answer will be negative the answer for positive $a$. This is because the integrand is an even function, but the order of the limits essentially flip in the negative case. For instance, if we integrated from $\frac{1}{3}$ to $3$, the answer from the negative version, which is integrating from $-\frac{1}{3}$ to $-3$, will be negative this (it'll be same absolute value, but has a negative sign because we are integrating with the more positive number, $-\frac{1}{3}$, in the \textsl{lower} bound instead of upper bound as usual. In other words, we are integrating in the `reverse order', from right to left instead of left to right as was done in the positive case here.).$

$\noindent Similarly, if we were integrating in `reverse order' in the positive case, it is easy to see that the negative version becomes integrating in the `correct' order. For example, if we integrated from $2$ to $\frac{1}{2}$ (a reverse order) in the positive case $\Big{(}$$a=\frac{1}{2}$$\Big{)}$, the negative version is integrating from $-2$ to $-\frac{1}{2}$, which is now the `correct' order, and so the sign flips. So using this notion, you should be able to get the right formula for negative $a$ too.$

$\noindent Note that since the integrand is positive everywhere in the interval of integration, as a check, the answer for $I(a)$ should be positive whenever the lower bound is less than the upper bound. So for positive $a$, whenever $\frac{1}{a}<a$ (i.e. if $a>1$), we should have our formula for $I(a)$ give a positive answer. If $0<a<1$, $\frac{1}{a}>a$ and we should have a negative answer from our formula. Note that $I(1)$ and $I(-1)$ should both be 0 (since the bounds are equal then), so another check is to make sure our formula obtained for $I(a)$ satisfy these.$

Paradoxica · May 19, 2016

SammyT123 said:
Also from the same booklet

$\int^{e}_{1/e}\frac{\arctan(x)dx}{x}$

I just tried IBP, and used the substitution $u=ln(1/x) $ for J$$

This gave me

$I = (\tan^{-1}{x}ln{x})^{e}_{1/e} - J \\ J= \int^{1}_{0}\frac{-udu}{e^u+e^{-u}}$

u=1/x, change variables back to x, add together and use inverse trig relations to simplify.

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Impossible question? (1 Viewer)

leehuan

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InteGrand

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Paradoxica

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