A Laplace transform

Calculus (Integrals, Series)
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Tolaso J Kos
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A Laplace transform

#1

Post by Tolaso J Kos »

The following integral was proposed by Seraphim

Prove that

$$\int_{0}^{\infty}\sin x \sin \left ( \sqrt{x} \right )e^{-x} \, {\rm d}x = \frac{\sqrt{\pi} \sqrt[4]{2}}{4\sqrt[8]{e}} \sin \left ( \frac{3\pi -1}{8} \right )$$

He has also proved that:

$$\int_{0}^{\infty}\sin x \sin \left ( \sqrt{x} \right )e^{-a x} \, {\rm d}x = \frac{\sqrt{\pi}}{2}\frac{1}{\sqrt[4]{\left ( a^2+1 \right )^3}} e^{- \frac{a}{4 \left ( a^2+1 \right )}} \sin \left ( \frac{3}{2} \arctan \frac{1}{a} - \frac{1}{4\left ( a^2+1 \right )} \right ) , \;\; a \geq 0$$
Imagination is much more important than knowledge.
r9m
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Re: A Laplace transform

#2

Post by r9m »

\begin{align*}I = \int_0^{\infty} e^{-ax}\sin x \sin \left(\sqrt{x}\right)\,dx &= \int_0^{\infty}xe^{-ax^2}\left(\cos \left(x-x^2\right)-\cos \left(x+x^2\right)\right)\,dx\\&=\mathfrak{Re} \int_0^{\infty}x\left(e^{-(a+i)x^2+ix}-e^{-(a+i)x^2-ix}\right)\,dx\\&= \mathfrak{Re} \int_{-\infty}^{\infty}xe^{-(a+i)x^2+ix}\,dx\\&= \mathfrak{Re} \left(e^{-\frac{1}{4(a+i)}}\int_{-\infty}^{\infty}xe^{-(a+i)\left(x^2-\frac{2i}{2(a+i)}x-\frac{1}{4(a+i)^2}\right)}\,dx\right)\\&= \mathfrak{Re} \left(e^{-\frac{1}{4(a+i)}}\int_{-\infty}^{\infty}xe^{-(a+i)\left(x-\frac{i}{2(a+i)}\right)^2}\,dx\right)\end{align*}

The function $f(z) = ze^{-(a+ib)z^2+iz}$ is a holomorphic function on the complex plane and $|f(z)| \to 0$ as $|z| \to \infty$, hence, we can shift the line of integration without worrying about residues.

Thus, \begin{align*}I &= \mathfrak{Re} \left(e^{-\frac{1}{4(a+i)}}\int_{-\infty}^{\infty}\left(x+\frac{i}{2(a+i)}\right)e^{-(a+i)x^2}\,dx\right)\\&= \mathfrak{Re} \left(\frac{ie^{-\frac{1}{4(a+i)}}}{2(a+i)}\int_{-\infty}^{\infty}e^{-(a+i)x^2}\,dx\right)\\&= \mathfrak{Re} \left(\frac{ie^{-\frac{1}{4(a+i)}}}{2(a+i)}\sqrt{\frac{\pi}{a+i}}\right)\\&= \frac{\sqrt{\pi}}{2(a^2+1)^{3/4}}e^{-\frac{a}{4(a^2+1)}}\cos \left(\frac{\pi}{2}+\frac{1}{4(a^2+1)}-\frac{3}{2}\arctan\frac{1}{a}\right)\\&= \frac{\sqrt{\pi}}{2(a^2+1)^{3/4}}e^{-\frac{a}{4(a^2+1)}}\sin \left(\frac{3}{2}\arctan\frac{1}{a}-\frac{1}{4(a^2+1)}\right)\end{align*}
galactus
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Re: A Laplace transform

#3

Post by galactus »

Clever and efficient, RD. :clap2:


Well, I scratched around with this one a little. I did not want to do the same exact thing as R. I do not know how Seraphim done it, so here goes:

$$\int_{0}^{\infty}e^{-ax}\sin(x)\sin(\sqrt{x})dx$$

Use identities for both sin terms, one the Taylor series for $\sin(\sqrt{x})$ and the other the complex form for sin(x):

$$\int_{0}^{\infty}\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)!}x^{k+1/2}e^{-ax}\left(\frac{e^{ix}-e^{-ix}}{2i}\right)dx$$

$$=\frac{1}{2i}\int_{0}^{\infty}\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)!}x^{k+1/2}(e^{x(i-a)}-e^{-x(i+a)})dx$$

Now, we have a kind of Gamma thing resulting from the integration:

$$=\frac{\sqrt{\pi}}{4i(i-a)^{3/2}}\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!4^{k}(i\pm a)^{k}}$$

Take the two separate $i\pm a$ terms and note the sum is merely that of e:

$$\sum_{k=0}^{\infty}\frac{\left(\frac{-1}{4(i-a)}\right)^{k}}{k!}=e^{\frac{-1}{4(i-a)}}$$

same for the other:

$$\sum_{k=0}^{\infty}\frac{\left(\frac{-1}{4(i+a)}\right)^{k}}{k!}=e^{\frac{-1}{4(i+a)}}$$

So, we ultimately have:

$$\frac{\sqrt{\pi}}{4i(i-a)^{3/2}}\left(e^{\frac{-1}{4(i-a)}}-e^{\frac{-1}{4(i+a)}}\right)$$

and this equals the required form.
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