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Problem:

$$\lim_{t \to 0}\int_{0}^{\frac{\pi}{2}}\frac{1 - \sin^t x}{t}dx$$

I have tried using L'Hopital Rule to calculate this but I could not go further. I also tried multiplying and dividing by $\cos x$ to force the by-parts rule but it was of no use either.

I would appreciate any kind of hint or idea towards the solution. As far as I am told, this problem isn't supposed to be very hard but I have given enough hours on it to come and ask here.

bigbang
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  • Try the lhopital rule to get the denominator as one in the integrand. – Тyma Gaidash Jul 05 '21 at 16:21
  • Do you mean that I should use lhopital on $$\frac{\int_0^{\pi/2} (1 - \sin^t x) dx}{t}$$ – bigbang Jul 05 '21 at 16:24
  • It seems that maybe:$$ \mathrm{\frac{\frac d{dt}1-sin^t(x)}{\frac d{dt}t}=-sin^t(x)(ln(sinx))\mathop {=}^{t=0} -ln(sin(x))}$$ – Тyma Gaidash Jul 05 '21 at 16:27
  • Okay, so we are taking the limit inside the integral. Oh, yes! I'm surprised how I applied a similar thing in a previous problem and missed it here. Thanks! – bigbang Jul 05 '21 at 16:32

2 Answers2

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Hint: $$\lim_{t \to 0} \frac{ \sin^t x - 1}{t} = \lim_{t \to 0} \frac{ e^{t\log(\sin x)} - 1}{t} =\log(sin(x))$$

which can give you convergence and thus you can interchange the limit and the integral.

Dosidis
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It seems that maybe:$$ \mathrm{\frac{\frac d{dt}1-sin^t(x)}{\frac d{dt}t}=-sin^t(x)(ln(sinx))\mathop {=}^{t=0} -ln(sin(x))}$$

Therefore:

$$\mathrm{I=-\int_0^\frac\pi 2 ln(sin(x))dx=\int_0^\frac \pi 2\sum_{n=1}^\infty\frac{(-1)^n(sin(x)-1)^n}{n}dx}$$

I will evaluate this perhaps with a series.

Here is the rest of the evaluation in this post.

You will get the final answer as: $$\mathrm{I= \lim_{t \to 0}\int_{0}^{\frac{\pi}{2}}\frac{1 - \sin^t x}{t}dx= \frac{\pi}2 ln(2)=1.088793045151…}$$

Тyma Gaidash
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  • The evaluation can be found here: https://math.stackexchange.com/a/2804742/194169. You can add this in your answer so that OP can accept it. – Dosidis Jul 05 '21 at 16:37
  • I dont think we need any series, there is simpler computation to it and I know that part. I will accept the answer as is. – bigbang Jul 05 '21 at 16:40
  • Yes, I see. There would have been a Hypergeometric function series to evaluate which I am not sure how to do. I will add it. Thanks. – Тyma Gaidash Jul 05 '21 at 16:42