Calculation of integral

[Tolaso J Kos]

Evaluate the integral:

$$\int_0^{\infty} t^{s-1} 3^{-t}\, {\rm d}t, \;\;\; \mathfrak{Re}(s)>0$$

[Papapetros Vaggelis]

Hello Tolaso

For \(\displaystyle{t>0}\) we have that

\(\displaystyle{t^{s-1}\,3^{-t}=t^{s-1}\,e^{\ln\,3^{-t}}=t^{s-1}\,e^{-\ln\,3\,t}}\) .

So, in order to calculate the integral \(\displaystyle{\int_{0}^{\infty}t^{s-1}\,3^{-t}\,\mathrm{d}t}\) , we make the substitution

\(\displaystyle{u=\ln\,3\,t}\) and thus: \(\displaystyle{u\in\left[0,+\infty\right)\,\,,t=\dfrac{u}{\ln\,3}\,\,,\mathrm{d}t=\dfrac{1}{\ln\,3}\,\mathrm{d}u}\)

so:

\(\displaystyle{\begin{aligned} \int_{0}^{\infty}t^{s-1}\,3^{-t}\,\mathrm{d}t&=\int_{0}^{\infty}\left(\dfrac{u^{s-1}}{(\ln\,3)^{s-1}}\right)\,e^{-u}\,\dfrac{1}{\ln\,3}\,\mathrm{d}u\\&=\dfrac{1}{(\ln\,3)^{s}}\,\int_{0}^{\infty}u^{s-1}\,e^{-u}\,\mathrm{d}u\\&\stackrel{Re(s)>0}{=}\dfrac{\Gamma(s)}{(\ln\,3)^{s}}\end{aligned}}\)

[Tolaso J Kos]

Hi there and thank you for your answer. Another way using brute force and heavily tools would be note that:

$$3^{-t}= \sum_{k=0}^{\infty}\frac{(-t)^k \log^k 3}{k!}$$

Then by applying Ramanujan master theorem the result follows since:

$$\int_{0}^{\infty}t^{s-1}3^{-t}\, {\rm d}t = \Gamma(s) \log^{-s} 3$$

Actually this is the Mellin Transformation of the function $f(t)=3^{-t}$ denoted as $\mathcal{M}\left ( 3^{-t} \right )$.

[Papapetros Vaggelis]

Tolaso, let me propose an integral in the same spirit.

Calculate the integral \(\displaystyle{\int_{0}^{\infty}3^{-4\,x^2}\,\mathrm{d}x}\) .

[Tolaso J Kos]

Tolaso, let me propose an integral in the same spirit.

Calculate the integral \(\displaystyle{\int_{0}^{\infty}3^{-4\,x^2}\,\mathrm{d}x}\) .

Hello Vaggelis,

here is a solution to your proposed int. using master theorem.

$$\begin{align*}
\int_{0}^{\infty}3^{-4x^2}\, {\rm d}x &\overset{u=4x^2}{=\! =\! =\!} \frac{1}{4}\int_{0}^{\infty}u^{-1/2} 3^{-u}\, {\rm d}u\\
&=\frac{1}{4}\mathcal{M}\left ( 3^{-u} \right ) \bigg|_{s=1/2} \\
&\overset{{\rm Master \; Theorem}}{=\! =\! =\! =\! =\! =\! =\! =\! =\!}\frac{1}{4}\cdot \frac{\Gamma (s)}{\log^s 3}\bigg|_{s=1/2} \\
&= \frac{1}{4}\cdot \frac{\Gamma \left ( \frac{1}{2} \right )}{\sqrt{\log 3}}= \frac{\sqrt{\pi}}{4\sqrt{\log 3}}
\end{align*}$$

and the calculations are over.

[Papapetros Vaggelis]

Thank you Tolaso for your answer. Here is another solution.

We write $\displaystyle{3^{-4\,x^2}=e^{-4\,x^2\,\ln\,3}}$ and by using the substitution $\displaystyle{u=4\,x^2\,\ln\,3}$ we get $\displaystyle{u\in\left[0,+\infty\right)\,\,,x=\dfrac{u^{1/2}}{2\,\sqrt{\ln\,3}}}$ and $\displaystyle{\mathrm{d}x=\dfrac{u^{-1/2}}{4\,\sqrt{\ln\,3}}\,\mathrm{d}u}$ and thus:

$$\displaystyle{\begin{aligned}\int_{0}^{\infty}3^{-4\,x^2}\,\mathrm{d}x&=\int_{0}^{\infty}e^{-u}\,\dfrac{u^{-1/2}}{4\,\sqrt{\ln\,3}}\,\mathrm{d}u\\&=\dfrac{1}{4\,\sqrt{\ln\,3}}\,\int_{0}^{\infty}u^{1/2-1}\,e^{-u}\,\mathrm{d}u\\&=\dfrac{\Gamma(1/2)}{4\,\sqrt{\ln\,3}}\\&=\dfrac{\sqrt{\pi}}{4\,\sqrt{\ln\,3}} \end{aligned}}$$