Finding Limits of Improper Integrals Using L'Hospital's Rule

[steven187]

hello all

well this is the first time Iv used latex it really took me along time to write, is it suppose to take that long or is there a better way of doing it?

anyway this is a small part of a bigger problem everytime i tried i always show that the limite does not exist, but when i chuck it into mathematica the limite does exist, can anybody help, its really awkward check it out

thanxs


$$\int_0^1\frac{\Pi\coth\Pi\times\x}{2}-\frac{1}{2\times\x}dx$$
$$\lim_{\epsilon \rightarrow 0} (\int_\epsilon^1\frac{\Pi\coth\Pi\times\x}{2}-\frac{1}{2\times\x}dx)$$
then eventually i get this
$$\lim_{\epsilon \rightarrow 0} (\frac{\log[\epsilon]}{2}+\frac{\log[\sinh\Pi]}{2}-\frac{\log[\sinh\Pi\epsilon]}{2})$$
but no matter how much i tried i cannot get it to equal
$$\frac{\log[\frac{\sinh\Pi}{\Pi}]}{2}$$

 

[TenaliRaman]

$$\int_0^1\frac{\pi\coth\pi x}{2}-\frac{1}{2x}dx$$
$$= [\frac{log(sinh\pi x)}{2} - \frac{log(x)}{2}]_0^1$$
$$= [\frac{log(\frac{sinh\pi x}{x})}{2}]_0^1$$
$$= \frac{log(sinh\pi)}{2} - \lim_{x \rightarrow 0}\frac{log(\frac{sinh\pi x}{x})}{2}$$
$$= \frac{log(sinh\pi)}{2} - \frac{log(\pi)}{2}$$

-- AI
[edit]
Please use \pi with small p and use x instead of \times when referring to variable x. \PI and \times look absolutely awful in your given expression

 

[steven187]

hello there

thanxs for the advice about latex i sure hope i will improve in the near future
anyway i don't really understand how you did this step

$$\lim_{x \rightarrow 0}\log(\frac{sinh\pi x}{x})=log(\pi)$$

thats exactly where I am getting concused am i missing something i am suppose to know?

thanxs

 

[TenaliRaman]

You can take lim inside log (since the log function is continuous)

-- AI

 

[steven187]

well this is what i have done so far but i don't see how that's going to get me to $$\pi$$, i tried plotting it on mathematica, it wouldn't plot

$$\lim_{x \rightarrow 0}\log(\frac{sinh\pi x}{x})$$
$$=\log(\lim_{x \rightarrow 0}\frac{sinh\pi x}{x})$$
$$=\log(\lim_{x \rightarrow 0}\frac{e^{\pi x}+e^{-\pi x}}{2x})$$

where am i going wrong?

 

[TenaliRaman]

$$sinh\pi x = \frac{e^{\pi x}-e^{-\pi x}}{2}$$
Apply L' Hospital.

-- AI

 

[steven187]

wow this L' Hospital stuff is great thanxs for the directions muchly appreciated

steven