# Find the limit and derivative of integral function.

$\psi_m(x)$ 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?

• As for your second part take a look at Lebiniz's Integral Rule – DanZimm Aug 23 '14 at 20:19
• Ah, so the derivative is $\frac{|x|^m sin(ln|x|)^m}{x}$? – flisk Aug 23 '14 at 20:59
• 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). – DanZimm Aug 23 '14 at 22:11
• Thanks! I understand, I did it with the chain rule. – flisk Aug 23 '14 at 22:16
• 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 – 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:

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

• But take in to account that $\log|x|<0$ when $x<1$. – GEdgar Aug 23 '14 at 20:19
• I don't get it, won't this just give me $\frac{-1}{m}<\psi_m(0)<\frac{1}{m}$? – flisk Aug 23 '14 at 20:51
• @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 $\;-\frac1m\;$ ... – Timbuc Aug 24 '14 at 19:32
• 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). – 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 $m=1,2,3,4,5$

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