# If $a,b,c$ are real numbers then the value of the integral

If $a,b,c$ are real numbers then compute the value of

$$\lim_{t\to0}\ln\left(\frac{1}{t}\int_0^t(1+a\sin(bx))^{c/x}dx\right)$$

My Thoughts:

Tried some substitutions. Like $1/x=t$.Doesn't help.There was a similarity to Riemann sums too. But it didn't turn out to be that.

• @RobertIsrael The question isn't framed by me.If that assumption makes it easier then probably yes..
– user220382
Commented Sep 18, 2016 at 20:20

This looks like the Fundamental Theorem of Calculus in disguise. If we let $f(x)=(1+a\sin(bx))^{c/x}$ and $F(x)$ equal the antiderivative of $f(x)$ and we move the limit inside the logarithm, we have $$\ln\left({\lim_{t\to0}\frac{F(t)-F(0)}{t-0}}\right)$$ which equals the natural logarithm of the derivative of $F(t)$ at $t=0$, which is just $\lim_{x\to0}\ln(f(x))$. This leaves us with $$\lim_{x\to0}\ln((1+a\sin(b*x))^{c/x})=\lim_{x\to0}\frac{c*\ln(1+a\sin(b*x))}{x}$$ From there, you should be able to use L'Hopital's rule to force an answer out of this question.

• One problem however: you still need to find the value of $f(0)$ =) Commented Sep 18, 2016 at 20:28
• Genius _/\_!!!! :-D
– user220382
Commented Sep 18, 2016 at 20:36
• Not really. The fundamental theorem of calculus tells you that the limit (without the logarithm) is the value of the integrand at $0$ provided the function is defined at $0$ in the first place (you also need continuous in a neighborhood of $0$, actually). You can't substitute this with a limit, without further checks: note that the integrand function is not defined at $0$, so we have an improper integral to begin with. Commented Sep 18, 2016 at 20:56

The integrand function is not defined at $0$, so before starting to compute the limit, we have to ensure the integral exists.

Since later we're interested in the the limit at $0$, we can just work in some unspecified neighborhood of $0$. In particular, we take it so that $1+a\sin(bx)>0$, which is possible because $\sin0=0$ and the sine is continuous.

If we prove that $$\lim_{x\to0}(1+a\sin(bx))^{c/x}$$ is finite, then we can extend the function by continuity and the integral will pose no problem of existence. Such a limit can be computed by first taking the logarithm, so let's study $$\lim_{x\to0}\ln\bigl((1+a\sin(bx))^{c/x}\bigr)= \lim_{x\to0}\frac{c\ln(1+a\sin(bx))}{x}$$ Well, this is the derivative at $0$ of the function $g(x)=c\ln(1+a\sin(bx))$; since $$g'(x)=\frac{abc\cos(bx)}{1+a\sin(bx)}$$ and $g'(0)=abc$, we have that $$\lim_{x\to0}(1+a\sin(bx))^{c/x}=e^{abc}$$ so our problem becomes to find $$\lim_{t\to0}\ln\left(\frac{1}{t}\int_0^t f(x)\,dx\right)$$ where (for $x$ in the above mentioned neighborhood of $0$) $$f(x)=\begin{cases} (1+a\sin(bx))^{c/x} & \text{if x\ne0}\\[4px] e^{abc} & \text{if x=0} \end{cases}$$ which is continuous.

Now apply continuity of $\ln$ and the fundamental theorem of calculus.