Evaluate $\int_{-\infty}^{\infty} e^{t \cdot x^{\alpha}} \cos(s \cdot x^{\alpha})dx$, expressing the answer in terms of $t$, $s$ and $\alpha$

Question

One of the maths groups I'm apart of on Facebook posts (usually) daily maths challenges. Typically they act as small brain teaser for when I wake up and I can solve them without much trouble. However, today's was more challenging:

(Note: blue indicates a variable and red indicates a constant)

After failing at being able to make much progress by hand, I attempted to plug the integral straight into WolframAlpha to help put me in the right direction. However, Wolfram wasn't able to solve the integral either. I'm aware that Wolfram isn't the be all end all of mathematics, but it's still quite rare for me to encounter a definite integral it can't do anything with.

I next booted up Desmos' Online Graphing Calculator and graphed $$y_{1}(m)=\displaystyle{\int_{0}^{m} (e^{t x^{\alpha}} \cos(s x^{\alpha}))} dx$$ $$\text{and}$$ $$y_{2}(m)=\displaystyle{\int_{-m}^{0} (e^{t x^{\alpha}} \cos(s x^{\alpha}))} dx$$ using a range of random values for the constants. Through this I was able to confidently come to the conclusion that, as $m \rightarrow \infty$, $|y_2| \rightarrow |y_1|$. Therefore;

$$\displaystyle{\int_{-\infty}^{\infty} (e^{t x^{\alpha}} \cos(s x^{\alpha}))} dx = 2 \displaystyle{\int_{0}^{\infty} (e^{t x^{\alpha}} \cos(s x^{\alpha}))} dx$$

Great, finally some progress. Returning to WolframAlpha I attempted to see if it is possible to determine the indefinite integral. To my surprise, it is!

$$\displaystyle{\int_{\text{ }}^{\text{ }} (e^{t x^{\alpha}} \cos(s x^{\alpha}))} dx = - \frac {1}{2a} \left( \Gamma \Big(\frac{1}{a}, (t+s \cdot i)x^a \Big) (t+s \cdot i)^{-\frac{1}{a}} + \Gamma \Big(\frac{1}{a}, (t-s \cdot i)x^a \Big) (t-s \cdot i)^{-\frac{1}{a}}\right)$$

Things are looking good now! I then substituted $x=0$ into the integrated function:

$$ - \frac {1}{2a} \left( \Gamma \Big(\frac{1}{a}, (t+s \cdot i)0^a \Big) (t+s \cdot i)^{-\frac{1}{a}} + \Gamma \Big(\frac{1}{a}, (t-s \cdot i)0^a \Big) (t-s \cdot i)^{-\frac{1}{a}}\right)$$ $$ = - \frac {1}{2a} \left( \Gamma \Big(\frac{1}{a}, 0 \Big) (t+s \cdot i)^{-\frac{1}{a}} + \Gamma \Big(\frac{1}{a}, 0 \Big) (t-s \cdot i)^{-\frac{1}{a}}\right)$$ $$=- \frac {1}{2a} \left( \Gamma \Big(\frac{1}{a}\Big) (t+s \cdot i)^{-\frac{1}{a}} + \Gamma \Big(\frac{1}{a}\Big) (t-s \cdot i)^{-\frac{1}{a}}\right) \small{\text{ }\textbf{[1]}}$$

Okay, so far so good. However... I hit a little snag when attempting to compute the integral function as $x \rightarrow \infty$ ...

$$ - \frac {1}{2a} \left( \Gamma \Big(\frac{1}{a}, (t+s \cdot i)\infty^a \Big) (t+s \cdot i)^{-\frac{1}{a}} + \Gamma \Big(\frac{1}{a}, (t-s \cdot i)\infty^a \Big) (t-s \cdot i)^{-\frac{1}{a}}\right)$$ $$= - \frac {1}{2a} \left( \Gamma \Big(\frac{1}{a}, (t+s \cdot i)\infty \Big) (t+s \cdot i)^{-\frac{1}{a}} + \Gamma \Big(\frac{1}{a}, (t-s \cdot i)\infty \Big) (t-s \cdot i)^{-\frac{1}{a}}\right)$$

And here's where the problem arises. Evaluating $\Gamma \Big(\frac{1}{a}, (t+s \cdot i)\infty \Big)$. For the moment, I'm going to finish solve the question with the assumption that $\Gamma \Big(\frac{1}{a}, (t+s \cdot i)\infty \Big) = \Gamma \Big(\frac{1}{a}, \infty \Big)$ but we'll come back to this later.

$$- \frac {1}{2a} \left( \Gamma \Big(\frac{1}{a}, (t+s \cdot i)\infty \Big) (t+s \cdot i)^{-\frac{1}{a}} + \Gamma \Big(\frac{1}{a}, (t-s \cdot i)\infty \Big) (t-s \cdot i)^{-\frac{1}{a}}\right)$$ $$=- \frac {1}{2a} \left( \Gamma \Big(\frac{1}{a}, \infty \Big) (t+s \cdot i)^{-\frac{1}{a}} + \Gamma \Big(\frac{1}{a}, \infty \Big) (t-s \cdot i)^{-\frac{1}{a}}\right)$$ $$=0 \small{\text{ }\textbf{[2]}}$$

From here, to finishing solving the original integral we simply need to evaluate $\small{\textbf{[2]}}-\small{\textbf{[1]}}$ and then multiply it by $2$ (remembering that so far we've only been evaluating $\displaystyle{\int_{0}^{\infty} (e^{t x^{\alpha}} \cos(s x^{\alpha}))} dx$), thus giving us a final answer of:

$$\displaystyle{\int_{-\infty}^{\infty} (e^{t x^{\alpha}} \cos(s x^{\alpha}))} dx = \frac {\Gamma \Big(\frac{1}{a}\Big)}{a} \left((t+s \cdot i)^{-\frac{1}{a}} + (t-s \cdot i)^{-\frac{1}{a}}\right)$$

While it seems that wrapped up nicely, there's still the problem of the earlier assumption. Im not certain that it was valid to assume that $\Gamma \Big(\frac{1}{a}, (t+s \cdot i)\infty \Big) = \Gamma \Big(\frac{1}{a}, \infty \Big)$. Attempting to evaluate $\displaystyle \lim_{x \to \infty} (t+s \cdot i)x^a$ wasn't any help either (however, $\displaystyle \lim_{x \to \infty} \Big| (t+s \cdot i)x^a \Big| = \infty$, which is what led me to make the assumption to begin with)

So, to simply put it, I want to know if the assumption $\Gamma \Big(\frac{1}{a}, (t+s \cdot i)\infty \Big) = \Gamma \Big(\frac{1}{a}, \infty \Big)$ is correct (or at least gave me the correct answer). If it's not, I'd very much appreciate it if someone could help explain the correct way to evaluate $\displaystyle{\int_{-\infty}^{\infty} (e^{t x^{\alpha}} \cos(s x^{\alpha}))} dx $ as I'm completely out of ideas at this point.

Answer 1

Your integral does not converge for all choices of $t,s,$ and $\alpha$. For example for $\alpha=1$, the integral diverges.

Let's assume $\alpha=2n$ with $n\in \{1,2,3,\dots\}$. Then you can write $\int_{-\infty}^\infty \dots = 2 \int_0^\infty \dots$, as you did (this does not work for any $\alpha$!).

Here is a sketch of a solution:

1. Substitute $y=x^{\alpha}$

2. Use $\cos(z)= \textrm{Re} \{\exp{(iz)}\}$ and move the $\textrm{Re}$ outside of the integral.

3. For your exponential under the integral use the series expansion $\exp(x)=\sum_{k=0}^\infty \frac{x^k}{k!}$ (for $x \in \mathbb{C}$), then use Ramanujan's Master theorem

4. Simplify your expression

If I made no mistakes the answer is

$$\frac{2}{\alpha} \Gamma \left(\frac{1}{\alpha} \right) \textrm{Re}\{(t-is)^{-1/\alpha} \}$$