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One useful formula evaluate some improper integrals is the following;

$$\boxed{\int_{0}^{\infty}\frac{f(t)}{t}\text{d}t=\int_{0}^{\infty}\mathcal{L}\{f(t)\}\text{d}s}$$

where $\mathcal{L}\{f(t)\}$ is the Laplace transform of the function $f(t)$.

I know an example of this which is

$$\int_{0}^{\infty}\frac{\sin(t)}{t}\text{d}t=\int_{0}^{\infty}\mathcal{L}\{\sin(t)\}\text{d}s=\int_{0}^{\infty}\frac{1}{s^2+1}\text{d}s=\tan^{-1}(s)|_{0}^{\infty}=\frac{\pi}{2}$$

Now I need to have some other examples to use this formula, I tried to construct (using inverse Laplace) but failed to do so.

Can you recommend me some easy examples as the one above?

Your help would be appreciated. Thanks!

Gary
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Hussain-Alqatari
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  • How about $f(t)=\sin(at)e^{-bt}$ for constants $a,,b>0$? – J.G. Sep 14 '22 at 11:05
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    $$\int_{0}^{\infty}\frac{\delta(t-\tau)}{t}{d}t=\int_{0}^{\infty}\mathcal{L}{\delta(t-\tau)} \ {d}s=\int_{0}^{\infty}e^{-\tau s }ds=\frac1{\tau}$$ – Quanto Sep 14 '22 at 12:18
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    Can I ask why you need to have some other examples? Are you presenting these to someone, are they to help you build an intuition of when you can apply this technique, or are you just curious? Knowing this might help others answer your question. – Christian E. Ramirez Sep 14 '22 at 22:11
  • @Quanto this does not seem true https://www.wolframalpha.com/input?i=integrate+diracdelta%28t-x%29%2Ft+dt+from+0+to+infty – Max0815 Sep 24 '22 at 05:43
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    @Max0815 - $\tau >0$ is implicitly assumed. – Quanto Sep 24 '22 at 15:55
  • Okay, I found another one. Technically... lol. We have $$f(x)=\frac{1}{x^{-2}+x^2}$$ with its laplace transform being MeijerG[{{1/4}, {}}, {{0, 1/4, 1/4, 1/2, 3/ 4}, {}}, s^4/256]/(4 Sqrt[2] \[Pi]^(3/2)) but mathematica can solve this meijer g integral quickly to be $\frac{\pi}4$. – Max0815 Sep 24 '22 at 20:06

2 Answers2

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If you read it backward, we can combine that identity with the Frullani integral to get the formula: $$ \int_{0}^{\infty} \alpha f(\alpha s) - \beta f(\beta s)\, \mathrm{d}s = \int_{0}^{\infty} \frac{\mathcal{L}^{-1}\left\{f\right\}\left( \frac{t}{\alpha}\right) - \mathcal{L}^{-1}\left\{f\right\}\left( \frac{t}{\beta}\right)}{t}\, \mathrm{d}t = \mathcal{L}^{-1}\left\{f\right\}\Big\vert_{t=0}^{t=\infty} \ln\left\vert\frac{\beta}{\alpha}\right\vert $$ where $\lim\limits_{t\to 0^+} \mathcal{L}^{-1}\left\{f(s)\right\}(t)$ and $\lim\limits_{t\to \infty} \mathcal{L}^{-1}\left\{f(s)\right\}(t)$ exist.

One example you where can use this is $$ \int_0^{\infty} \arctan\left( \frac{1}{x\pm 1}\right) - \frac{1}{\xi}\arctan\left(\frac{\xi}{x \pm \xi}\right)\, \mathrm{d}x = \ln|\xi| $$ where two sign choices are independent of each other. This integral looks rather complicated considering the fact that the integrand can be discontinuous and that splitting the integral results into two divergent integrals. However, if you know that $$\mathcal{L}\left\{\frac{\sin(t)}{t}e^{-t} \right\} = \int_0^{\infty}\frac{\sin(t)e^{-t(s+1)}}{t}\, \mathrm{d}t= \int_0^{\infty}\int_0^{\infty} e^{-t(s+1 +x)}\sin(t)\,\mathrm{d}t\, \mathrm{d}x=\arctan\left(\frac{1}{s+1}\right) $$ then an immediate application of the formula with $\alpha = \pm 1$ and $\beta = \pm\frac{1}{\xi}$ gives the result.

Robert Lee
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Inn this posting I describe a class of functions for which the formula stated by the OP holds.

Suppose $\int^\infty_0\frac{f(t)}{t}\,dt$ converges, that is, for $F(t)=\int^t_0 \frac{f(x)}{x}\,dx$, $\lim_{t\rightarrow\infty}F(t)$ exists

For $s>0$, the function $$\Phi(s)=\int^\infty_0\frac{f(t)}{t}e^{-st}\,dt$$ is well defined (the integral converges for all $s>0$), and $$\lim_{s\rightarrow0}\Phi(s)=\int^\infty_0\frac{f(t)}{t}\,dt$$ (see here for example).

If we assume that for each $s>0$, the integral $\Phi(s)$ converges itself in the sense of Lebesgue, then we have that $\lim_{s\rightarrow\infty}\Phi(s)=0$, and that $$\Phi'(s)=-\int^\infty_0f(t)e^{-st}\,dt=-Lf(s)$$ Hence $$-\Phi(s)+\Phi(0+)=\int^s_0Lf(y)\,dy$$

Letting $s\rightarrow\infty$ yields

$$\Phi(0+)=\int^\infty_0\frac{f(t)}{t}\,dt=\int^\infty_0 Lf(t)\,dt$$

Some examples:

  • $f(t)=e^{-cs}\sin t$ for $c>0$
  • $f(t)=\frac{1-e^{-t^2}}{t}$
Mittens
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