Find the value of this definite integral in terms of t, s and alpha

[Saracen Rue]

TL;DR Summary

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

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. I am 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.

 

[anuttarasammyak]

How about using complex integral ? Given I is

$$I= Re \int_{-\infty}^{+\infty} e^{-ux^{\alpha}} dx = Re\ u^{-\frac{1}{\alpha}}\int_C e^{-z^{\alpha}}dz$$ where
$$u=t+is$$
Oh, I can not go further.

 

[Delta2]

I am not sure I understand all your analysis correctly, for example i can see that for t=s=a=1 the integral
$$\int_{-\infty}^0 e^{-x}\cos x dx=+\infty$$
while the integral
$$\int_{0}^{+\infty} e^{-x}\cos x dx=0.5$$
so your very starting point
$$\displaystyle{\int_{-\infty}^{\infty} (e^{t x^{\alpha}} \cos(s x^{\alpha}))} dx = 2 \displaystyle{\int_{0}^{\infty} (e^{t x^{\alpha}} \cos(s x^{\alpha}))} dx$$
doesn't seem to be valid , at least not for all the combinations of constant values t,s,a.
I think that the integral generally does not converge, for example for s=t ( or s=-t) and a>0 I can see that the integral does not converge.

 

[Saracen Rue]

I am not sure I understand all your analysis correctly, for example i can see that for t=s=a=1 the integral
$$\int_{-\infty}^0 e^{-x}\cos x dx=+\infty$$
while the integral
$$\int_{0}^{+\infty} e^{-x}\cos x dx=0.5$$
so your very starting point
$$\displaystyle{\int_{-\infty}^{\infty} (e^{t x^{\alpha}} \cos(s x^{\alpha}))} dx = 2 \displaystyle{\int_{0}^{\infty} (e^{t x^{\alpha}} \cos(s x^{\alpha}))} dx$$
doesn't seem to be valid , at least not for all the combinations of constant values t,s,a.
I think that the integral generally does not converge, for example for s=t ( or s=-t) and a>0 I can see that the integral does not converge.

Thanks for the reply. I did the testing regarding this part using Desmos, which said that the integral should be converging to $1$ for the constants all equalling $1$.

In fact, Desmos indicates that the integral converges so long as all the constants are are greater than $0$:

 

[Delta2]

Desmos seems right, however you do a "tricky "mistake in the integral setup of $y_4$ or$y_2$ or you ask for the wrong value,
for example since you setup $$y_4=\int_{-x}^0 e^{-m}\cos m dm$$ and then you ask it for $$y_4(-10000000)$$ Desmos calculates $$\int_{-(-1000000)}^0 e^{-m}\cos m dm$$ and not $$\int_{-1000000}^0 e^{-m}\cos m dm$$ as I assume you wanted it to calculate.

 

[anuttarasammyak]

How about using complex integral ? Given I is

$$I= Re \int_{-\infty}^{+\infty} e^{-ux^{\alpha}} dx = Re\ u^{-\frac{1}{\alpha}}\int_C e^{-z^{\alpha}}dz$$ where
$$u=t+is$$
Oh, I can not go further.

With suggestion of post #3 by Delta2, let say $\alpha=2n$ , n is a positive integer, so RHS integral which has no residue could equal to
$$2\int_0^\infty e^{-x^{2n}}dx=\frac{1}{n}\Gamma(\frac{1}{2n})$$
with expectation that integral at R$\rightarrow \infty$ make no contribution,
and
$$Re\ u^{-\frac{1}{\alpha}}=(t^2+s^2)^{-\frac{1}{4n}}Re\ e^{-i\frac{\theta}{2n}}=(t^2+s^2)^{-\frac{1}{4n}}\cos\frac{\theta}{2n}$$ where
$$\theta=tan^{-1} \frac{s}{t}$$