$\psi_m(x)$ is defined as $$\int_0^{\ln|x|}e^{mt}\sin(t)^m\mathop{dt}$$ with $m$ a natural number greater then zero. Now the question is, does $\lim\limits_{x\to 0}\psi_m(x)$ exist. I've tried using the squeeze theorem using $-1\leq \sin(t)^m\leq 1$ and this resulted in $\frac{-1}{m}\leq\psi_m(x)\leq\frac{1}{m}$, which isn't useful.

Another part of the question was to determine the derivative. But I have no idea how, since the variable $x$, is in the limit of integration. I tried doing integration by parts, in hope of finding recursion, but this didn't really work (integrated by parts twice).

Any ideas how to solve this?

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    $\begingroup$ As for your second part take a look at Lebiniz's Integral Rule $\endgroup$ – DanZimm Aug 23 '14 at 20:19
  • $\begingroup$ Ah, so the derivative is $\frac{|x|^m sin(ln|x|)^m}{x}$? $\endgroup$ – flisk Aug 23 '14 at 20:59
  • $\begingroup$ I believe so, another way of looking at this is $f(y) = \int_0^{y} e^{mt} \sin(t)^m \, dt$ and then note that you're trying to find $\frac{d}{dx} f(\ln \lvert x \rvert)$ and then of course you have the chain rule (this is personally how I remember Lebiniz's rule). $\endgroup$ – DanZimm Aug 23 '14 at 22:11
  • $\begingroup$ Thanks! I understand, I did it with the chain rule. $\endgroup$ – flisk Aug 23 '14 at 22:16
  • $\begingroup$ I think if you change the $\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$ and then use the binomial theorem you'll get a pretty decent simplification I think. I would write out a full response but I got a flu shot yesterday and I'm pretty sure I have horrid immune system because I feel pretty sick today :P $\endgroup$ – DanZimm Aug 23 '14 at 22:20

Since $e^{mt}\sin(t)^m$ is a $L^1(\mathbb{R}^-)$ function, we just have to compute: $$I_m=\int_{0}^{+\infty}\sin(t)^m e^{-mt}dt.$$ Notice that we have: $$\int_{0}^{+\infty}e^{i\eta t}\,e^{-mt}\,dt=\frac{1}{m-i\eta}\tag{1}$$ while the binomial theorem gives: $$\sin(t)^m = \frac{1}{(2i)^m}\sum_{j=0}^{m}\binom{m}{j}(-1)^j e^{(m-2j)it}\tag{2}$$ so it follows that: $$ I_m = \frac{1}{(2i)^m}\sum_{j=0}^{m}\binom{m}{j}\frac{(-1)^j}{m-(m-2j)i}$$ or:

$$ I_{2m} = \frac{1}{4^m}\left(\frac{1}{2m}\binom{2m}{m}+(-1)^m\sum_{j=0}^{m-1}\binom{2m}{j}\frac{(-1)^j m}{m^2+(m-j)^2}\right),$$

$$ I_{2m+1} = \frac{(-1)^m}{2\cdot 4^m}\sum_{j=0}^{m}\binom{2m+1}{j}\frac{(-1)^j(2m-2j+1)}{j^2+(2m+1-j)^2}.\tag{3}$$

So for any $m\in\mathbb{N}$ we have that $\lim_{x\to 0^+}\psi_m(x)$ exists and equals a rational number: $$ \lim_{x\to 0^+}\psi_m(x) = (-1)^m\, I_m.$$


As a hint:

$$\left|\int\limits_0^{\log|x|}e^{mt}\sin^mt\,dt\right|\le\int\limits_0^{\log |x|}e^{mt}dt=\left.\frac1me^{mt}\right|_0^{\log|x|}=\frac1m\left(|x|^m-1\right)\xrightarrow[x\to 0]{}...?$$

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    $\begingroup$ But take in to account that $\log|x|<0$ when $x<1$. $\endgroup$ – GEdgar Aug 23 '14 at 20:19
  • $\begingroup$ I don't get it, won't this just give me $\frac{-1}{m}<\psi_m(0)<\frac{1}{m}$? $\endgroup$ – flisk Aug 23 '14 at 20:51
  • $\begingroup$ @GEdgar, I don't understand: what does that have to do with my hint? If $\;|x|<1\,,\,x\neq 0\;$ then $\;\log|x|<0\;$ , and in the integral the upper limit is less than the lower one...but the result remains and the limit's still $\;-\frac1m\;$ ... $\endgroup$ – Timbuc Aug 24 '14 at 19:32
  • $\begingroup$ No, then it becomes $\frac{1}{m}$, and this is what I already found. The sign switches when you switch the integral limits. It's impossible to get $\frac{-1}{m}$, because then the absolute value of the integral would become less then a negative number (which is impossible). $\endgroup$ – flisk Aug 24 '14 at 20:46

Basically you are looking for evaluating the integral

$$I = \int_{0}^{-\infty} e^{mt}\sin(t)^m\, dt. $$

A possible closed form is

$$I= {\frac{\left( -1/2-i/2 \right) \Gamma( m ) \Gamma\left( -m/2 + 1-im/2\right) }{( -2\,i )^{m}\,\Gamma( -im/2+m/2+1 ) }}, \quad i=\sqrt{-1}. $$

Here some special values for $m=1,2,3,4,5$

$$ \left\{ \frac{1}{2},-\frac{1}{8},\frac{1}{30},-{\frac {3}{320}},{\frac {3}{1105}} \right\} $$


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