# I Integral e^(ikx^3)

1. Dec 5, 2017

### spacetimedude

Show that $$\int_0^\infty dx\exp(ikx^3) , k>0$$ may be written as integral from 0 to $\infty$ along the line $arg(z) = \frac{\pi}{6}$.

I'd appreciate it if you can help me how to approach this problem. My initial impression was to expand the integrand out
$$\sum^{\infty}_{n=0}\frac{(ikx^3)^n}{n!}$$
but did not how to obtain the $arg(z)$ condition. I plugged the integral in wolframalpha and gave me an expression with a Gamma function, which the lecture has covered but I'm not sure how to apply here.

Thanks

2. Dec 5, 2017

### Staff: Mentor

Which expression in terms of the gamma-function do you get from the substitution $t = -ikx^3\,$?

3. Dec 5, 2017

### spacetimedude

Wolfram says $$\frac{\Gamma(\frac{4}{3})}{\sqrt[3]{-ik}}.$$
I have been told that using $i=\exp(i\frac{\pi}{2})$ helps but I can only get as far as writing the integrand as $\exp(\exp(i\frac{\pi}{2})kx^3))$ and not sure what to do after.

4. Dec 5, 2017

### Staff: Mentor

That wasn't my question. I've asked what do you get, not WolframAlpha. The substitution of $i$ might help to properly calculate the powers of $i$ but I'd rather write $k=ic$ to begin. In any case one has to be cautious with the handling of the purely imaginary constants.

The substitution $t \mapsto -ikx^3$ allows to write $x$ as power of $t$ as well as the integral in terms of the gamma function.

5. Dec 5, 2017

### Delta²

I think the exercise asks not to compute the value of the integral but to express it as a contour integral with the contour being the line $arg(z)=\frac{\pi}{6}$.

6. Dec 5, 2017

### spacetimedude

Yes, the expression is what I am looking for.
I am solving one of the past exam papers for my university and it is the first part of the question.

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7. Dec 5, 2017

### spacetimedude

Okay, so I am looking for the expression of the contour integral as Delta said. I allowed $i=\exp(i\frac{\pi}{2})$ and extend the integral to z space. Then let $z=r\exp(i\theta)$, yielding
$$\int _C dz e^{{kr^2e^{i(\pi/2+3\theta)}}}.$$

Do you think this is correct so far? Maybe $z=r\exp(i\theta)$ is unnecessary.

Last edited: Dec 5, 2017
8. Dec 5, 2017

### spacetimedude

What I have now is:
Choose $z=re^{i\pi/6}$ so $dz=e^{i\pi/6}dr$. Plug in the integral ->
$$e^{i\pi/6}\int^\infty_{-\infty} e^{ikr^3e^{i\pi/2}}$$ and note $e^{i\pi/2}=i$
Finally, the integral becomes $$e^{i\pi/6}\int^\infty_{-\infty} e^{-kr^3}.$$
Does this seem right?

9. Dec 6, 2017

### Delta²

The limits of integration should be from 0 to $+\infty$ (the contour is the positive part of the line $z=re^{i\pi/6}$.

However in order to be rigorous and clear you have to provide a function $f:\mathbb{C}->\mathbb{C}$ such that
$\int_0^{\infty}f(z(r))z'(r)dr=\int_0^{\infty}e^{ikr^3}dr$