How to calculate the following asymptotic approximation without using integration by part:
$$I(x) = \int_0^1 t^x \cos^n(\pi t)dt, \ \ \ \text{as} \ \ x \rightarrow \infty$$
Note: $n$ is an integer.
I try to do the following:
$$I(x) = \int_0^1 \ln e^{t^x} \cos^n(\pi t)dt, \ \ \ \text{as} \ \ x \rightarrow \infty$$
And then I have no idea to go further; how to transform it to use Gaussian integral? Please advise, thanks!
We may try this path.
First of all exponentiate the cosine:
$$\cos^n(\pi t) = \left(\frac{e^{i\pi t} + e^{-i\pi t}}{2}\right)^n = \frac{1}{2^n}\left(e^{i\pi t} + e^{-i\pi t}\right)^n$$
Then use the Binomial Series to expand the power
$$\left(e^{i\pi t} + e^{-i\pi t}\right)^n = \sum_{k = 0}^n \binom{n}{k}e^{ik\pi t} e^{-i\pi t(n-k)} = \sum_{k = 0}^n \binom{n}{k} e^{i\pi t(2k - n)}$$
Put all together and you have
$$\frac{1}{2^n} \sum_{k = 0}^n \binom{n}{k} \int_0^1 t^x e^{i\pi t(2k - n)}\ \text{d}t$$
By calling $\alpha = \pi(2k-n)$ we recognize a well known integral:
$$\int t^x e^{i\alpha t}\ \text{d}t = \frac{e^i t^x (-\alpha t)^{-x} \Gamma (x+1,-\alpha t)}{\alpha}$$
Integrating it from $0$ to $1$ yields
$$e^i \pi ^{-x-1} (n-2 k)^{-x-1} (\Gamma (x+1)-\Gamma (x+1,(n-2 k) \pi ))$$
Putting all together:
$$e^i\frac{1}{2^n}\frac{1}{\pi ^{x+1}} \sum_{k = 0}^n \binom{n}{k} \frac{1}{(n-2 k)^{x+1}} (\Gamma (x+1)-\Gamma (x+1,(n-2 k) \pi ))$$
By some knowledge about asymptotic expansion, you can show that around infinity:
$$\Gamma (x+1)-\Gamma (x+1,(n-2 k) \pi ) \sim \frac{\left[\pi (n - 2k)\right]^{x+1}}{x}e^{2k\pi - n\pi}$$
So using this we arrive at
$$e^i\frac{1}{2^n} \frac{e^{-\pi n}}{x}\sum_{k = 0}^n \binom{n}{k} e^{2kn} (n-2k)^{x+1}$$