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$$ \int^{+\infty}_0\frac{e^{-t}} {\sqrt t} \, dt$$

I have computed it getting $$ \frac{2e}{3} \therefore \text{ it converges.}$$ using integration by parts letting $$u = \frac 1 {\sqrt t} $$ and $$ dV= e^{-t} \, dt$$ Is it the right way to do it ?

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  • $\begingroup$ Do you know of the gamma function? $\endgroup$
    – TheSimpliFire
    Aug 5, 2018 at 14:48
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    $\begingroup$ @TheSimpliFire no I don't $\endgroup$
    – M12567
    Aug 5, 2018 at 14:49
  • $\begingroup$ @Mayar Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… $\endgroup$
    – user
    Sep 6, 2018 at 23:08

5 Answers 5

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If you simply want to decide convergence and you don't notice the trick in Henry Lee's answer, you can look at the following. Near zero, $$ \int_0^1 \frac{e^{-t}}{\sqrt t}\,dt\leq \int_0^1\frac1{\sqrt t}\,dt, $$ which is convergent.

At infinity, $$ \int_1^\infty \frac{e^{-t}}{\sqrt t}\,dt\leq\int_1^\infty e^{-t}\,dt<\infty. $$ So the integral converges.

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  • $\begingroup$ It is exactly the same idea I had. I let also my answer since I've suggested e different criteria for convergence. $\endgroup$
    – user
    Aug 6, 2018 at 8:12
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$$I=\int_0^\infty\frac{e^{-t}}{\sqrt{t}}\,dt$$ $u=\sqrt{t}\,,dt=2\sqrt{t}du$ $$I=2\int_0^\infty e^{-u^2}\,du=\sqrt{\pi}$$ As it is a standard integral

EDIT: $$I=2\int_0^\infty e^{-x^2}\,dx$$ then $$I^2=4\left(\int_0^\infty e^{-x^2}\,dx\right)^2=4\left(\int_0^\infty e^{-x^2}\right)\left(\int_0^\infty e^{-y^2} \, dy\right) = 4\int_0^\infty \int_0^\infty e^{-(x^2+y^2)} \, dx \, dy$$ now we can use polar coordinates to simplify this. $x^2+y^2=r^2\,$ and $dA=dx\,dy=r\,dr\,d\theta$ so our integral becomes: $$I^2=4\int_0^{\frac{\pi}{2}} \int_0^\infty e^{-r^2}r\,dr\,d\theta$$ now $u=-r^2$ so $\frac{du}{dr}=-2r\, \therefore\,dr=\frac{du}{-2r}$ and the integral becomes: $$I^2=-2\int_0^{\frac{\pi}{2}} \int_0^{-\infty}e^u \, du \, d\theta = 2\int_0^{\frac{\pi}{2}} \int_{-\infty}^0 e^u\,du\,d\theta = 2\int_0^{\frac{\pi}{2}} \left[e^u\right]_{-\infty}^0 \, d\theta = 2\int_0^{\frac{\pi}{2}} \, d\theta =2\cdot\frac{\pi}{2}=\pi$$ so if $I^2=\pi$ then $I=\sqrt{\pi}$

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  • $\begingroup$ This can also be done using the gamma function then linking to factorial but this is a well known result so thought this way is easier $\endgroup$
    – Henry Lee
    Aug 5, 2018 at 14:51
  • $\begingroup$ Yes this is easier , could you show me how you computed $$ \ sqrt(\ pi)$$ though , I know that by getting$$ \int {v dU}$$ its gonna be $$ = \ 1/2 .(I) $$ $\endgroup$
    – M12567
    Aug 5, 2018 at 19:28
  • $\begingroup$ I don't really understand your notation, but I will add to my answer to show why this is the case $\endgroup$
    – Henry Lee
    Aug 5, 2018 at 19:30
  • $\begingroup$ Okay thank you . $\endgroup$
    – M12567
    Aug 5, 2018 at 19:31
  • $\begingroup$ I have now added a proof. This uses polar coordinates which you may not be familiar with, and is commonly known as the Gaussian integral $\endgroup$
    – Henry Lee
    Aug 5, 2018 at 19:44
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$$\int_0^{+\infty}\frac{e^{-t}}{\sqrt{t}}dt=\int_{0}^{1}\frac{e^{-t}}{\sqrt{t}} \, dt + \int_1^{+\infty}\frac{e^{-t}}{\sqrt{t}} \, dt=A+B$$ for $A$ when $e^{-t}\sim1$ then $$A=\int_0^1 \frac{e^{-t}}{\sqrt{t}}\,dt\sim\int_0^1 \frac{1}{\sqrt{t}} \, dt < \infty$$ for $B$ when $t\geq1$ then $$B=\int_{1}^{+\infty}\frac{e^{-t}}{\sqrt{t}}\,dt\leq\int_1^{+\infty}e^{-t} \, dt < \infty$$

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  • $\begingroup$ That's also exactly my idea to proceed. Maybe for the integral A you should recall that we need to refer to limit comparison test to prove convergence. $\endgroup$
    – user
    Aug 6, 2018 at 8:33
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As Henry Lee's answer mentions, you get $\sqrt\pi,$ so $2e/3$ is wrong. But if the function is everywhere positive and you computed it correctly and get a finite number rather than $+\infty,$ then it converges. If it's not everywhere positive, then the meaning of convergence may become problematic. For example, $$ \lim_{a\,\to\,+\infty}\int_0^a \frac{\sin x} x \, dx = \frac \pi 2 $$ and that is a sort of convergence, but $$ \int\limits_{\{\,x \,:\, (\sin x)/x \, \ge\,0\}} \frac{\sin x} x\,dx = +\infty \quad\text{ and } \quad \int\limits_{\{\,x \,:\, (\sin x)/x \, < \,0\}} \frac{\sin x} x\,dx = -\infty $$ and some questions arise about when one should consider the thing convergent.

But one thing to bear in mind is that the question of whether something converges or not is often simpler than the question of what it converges to. Thus $$ 0 < \int_0^1 \frac{e^{-t}}{\sqrt t} \, dt \le \int_0^1 \frac 1 {\sqrt t} \, dt < +\infty $$ and $$ 0 < \int_1^\infty \frac{e^{-t}}{\sqrt t}\,dt \le \int_1^\infty e^{-t} \, dx = e^{-1} < +\infty $$ so what you have converges.

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If we can calculate a finite value for the integral of course the integral converges but if our goal is simply to check for the convergence we don’t need to calculate it explicitly but we can proceed for example by some quick convergence test.

For example, in this case we have that

$$ \int^{+\infty}_0\frac{e^{-t}} {\sqrt t} \, dt=\int^{1}_0\frac{e^{-t}} {\sqrt t} \, dt+ \int^{+\infty}_1\frac{e^{-t}} {\sqrt t} \, dt$$

and as $t\to 0^+$

$$\frac{e^{-t}} {\sqrt t} \sim \frac{1} {\sqrt t}$$

and as $t\to \infty$

$$\frac{e^{-t}} {\sqrt t} \sim \frac{1} {e^t}$$

therefore both integral converge by limit comparison test and therefore the given integral converges.

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    $\begingroup$ you are late ;) $\endgroup$
    – Nosrati
    Aug 6, 2018 at 8:07
  • $\begingroup$ @user108128 Sometimes it is good, so I can think to a better answer ;) $\endgroup$
    – user
    Aug 6, 2018 at 8:08
  • $\begingroup$ @user108128 I see now that my answer is very similar to that given by Martin, I let mine since I gave a suggestion by limit comparison test which avoid to use inequalities. $\endgroup$
    – user
    Aug 6, 2018 at 8:11

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