How to determine if an improper integral converges or diverges?
$begingroup$
Let's say we have
$int_{0}^{infty}frac{e^{-x}}{sqrt{x}}dx$.
I'd like to find the antiderivative and then analyze the integral going from 0 to N as N goes to infinity. However, I don't know how to find this particular primitive, and the online calculators do something that we haven't gone over in class (they use the Gauss error function), which leads me to suspect that there's another way of doing this.
Does anyone have an idea? Any help is appreciated, thank you.
integration riemann-integration
asked Dec 4 '18 at 23:37
user494405user494405
37919
$endgroup$
add a comment |
$begingroup$
Let's say we have
$int_{0}^{infty}frac{e^{-x}}{sqrt{x}}dx$.
I'd like to find the antiderivative and then analyze the integral going from 0 to N as N goes to infinity. However, I don't know how to find this particular primitive, and the online calculators do something that we haven't gone over in class (they use the Gauss error function), which leads me to suspect that there's another way of doing this.
Does anyone have an idea? Any help is appreciated, thank you.
integration riemann-integration
asked Dec 4 '18 at 23:37
user494405user494405
37919
$endgroup$
1
$begingroup$
a) Do you really want to mix $x$ and $t$? b) If you just want to know some estimates, then estimate this is reasonably close to $int_0^1 t^{-1/2} , dt$.
$endgroup$
– T. Bongers
Dec 4 '18 at 23:38
1
$begingroup$
You have an $e^{-x}$ in your integrand. Is that $x$ meant to be a $t$? If not, the integrand is much more trivial to deal with since it'd be a constant w.r.t. $t$.
$endgroup$
– Eevee Trainer
Dec 4 '18 at 23:38
$begingroup$
@T.Bongers Apologies, had a brain fart. Yes, I meant to have everything be the same variable; fixed now.
$endgroup$
– user494405
Dec 4 '18 at 23:40
add a comment |
$begingroup$
Let's say we have
$int_{0}^{infty}frac{e^{-x}}{sqrt{x}}dx$.
I'd like to find the antiderivative and then analyze the integral going from 0 to N as N goes to infinity. However, I don't know how to find this particular primitive, and the online calculators do something that we haven't gone over in class (they use the Gauss error function), which leads me to suspect that there's another way of doing this.
Does anyone have an idea? Any help is appreciated, thank you.
integration riemann-integration
asked Dec 4 '18 at 23:37
user494405user494405
37919
$endgroup$
Let's say we have
$int_{0}^{infty}frac{e^{-x}}{sqrt{x}}dx$.
I'd like to find the antiderivative and then analyze the integral going from 0 to N as N goes to infinity. However, I don't know how to find this particular primitive, and the online calculators do something that we haven't gone over in class (they use the Gauss error function), which leads me to suspect that there's another way of doing this.
Does anyone have an idea? Any help is appreciated, thank you.
integration riemann-integration
integration riemann-integration
asked Dec 4 '18 at 23:37
user494405user494405
37919
asked Dec 4 '18 at 23:37
user494405user494405
37919
edited Dec 4 '18 at 23:39
user494405
asked Dec 4 '18 at 23:37
user494405user494405
37919
asked Dec 4 '18 at 23:37
user494405user494405
37919
asked Dec 4 '18 at 23:37
user494405user494405
37919
37919
1
$begingroup$
a) Do you really want to mix $x$ and $t$? b) If you just want to know some estimates, then estimate this is reasonably close to $int_0^1 t^{-1/2} , dt$.
$endgroup$
– T. Bongers
Dec 4 '18 at 23:38
1
$begingroup$
You have an $e^{-x}$ in your integrand. Is that $x$ meant to be a $t$? If not, the integrand is much more trivial to deal with since it'd be a constant w.r.t. $t$.
$endgroup$
– Eevee Trainer
Dec 4 '18 at 23:38
$begingroup$
@T.Bongers Apologies, had a brain fart. Yes, I meant to have everything be the same variable; fixed now.
$endgroup$
– user494405
Dec 4 '18 at 23:40
add a comment |
1
$begingroup$
a) Do you really want to mix $x$ and $t$? b) If you just want to know some estimates, then estimate this is reasonably close to $int_0^1 t^{-1/2} , dt$.
$endgroup$
– T. Bongers
Dec 4 '18 at 23:38
1
$begingroup$
You have an $e^{-x}$ in your integrand. Is that $x$ meant to be a $t$? If not, the integrand is much more trivial to deal with since it'd be a constant w.r.t. $t$.
$endgroup$
– Eevee Trainer
Dec 4 '18 at 23:38
$begingroup$
@T.Bongers Apologies, had a brain fart. Yes, I meant to have everything be the same variable; fixed now.
$endgroup$
– user494405
Dec 4 '18 at 23:40
1
1
$begingroup$
a) Do you really want to mix $x$ and $t$? b) If you just want to know some estimates, then estimate this is reasonably close to $int_0^1 t^{-1/2} , dt$.
$endgroup$
– T. Bongers
Dec 4 '18 at 23:38
$begingroup$
a) Do you really want to mix $x$ and $t$? b) If you just want to know some estimates, then estimate this is reasonably close to $int_0^1 t^{-1/2} , dt$.
$endgroup$
– T. Bongers
Dec 4 '18 at 23:38
1
1
$begingroup$
You have an $e^{-x}$ in your integrand. Is that $x$ meant to be a $t$? If not, the integrand is much more trivial to deal with since it'd be a constant w.r.t. $t$.
$endgroup$
– Eevee Trainer
Dec 4 '18 at 23:38
$begingroup$
You have an $e^{-x}$ in your integrand. Is that $x$ meant to be a $t$? If not, the integrand is much more trivial to deal with since it'd be a constant w.r.t. $t$.
$endgroup$
– Eevee Trainer
Dec 4 '18 at 23:38
$begingroup$
@T.Bongers Apologies, had a brain fart. Yes, I meant to have everything be the same variable; fixed now.
$endgroup$
– user494405
Dec 4 '18 at 23:40
$begingroup$
@T.Bongers Apologies, had a brain fart. Yes, I meant to have everything be the same variable; fixed now.
$endgroup$
– user494405
Dec 4 '18 at 23:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Finding an antiderivative is going to be difficult. Seeing if it converges will not,
$$
int_0^infty frac{e^{-x}}{sqrt{x}}mathrm dx=
int_0^1 frac{e^{-x}}{sqrt{x}}mathrm dx+int_1^infty frac{e^{-x}}{sqrt{x}}mathrm dx
$$
Now, for $0leq xleq 1$, we have
$$
frac{e^{-x}}{sqrt{x}}leq frac{1}{sqrt{x}}
$$
and for $1leq x$, we have
$$
frac{e^{-x}}{sqrt{x}}leq e^{-x}
$$
Alternatively, enforce the substitution $u=sqrt{x}$, then your integral is
$$
int_0^infty frac{e^{-x}}{sqrt{x}}mathrm dx=
2int_0^infty e^{-u^2}mathrm du=sqrt{pi}
$$
answered Dec 4 '18 at 23:42
qbertqbert
22.1k32560
$endgroup$
$begingroup$
Thanks for your reponse. Sorry to be dense, but I don't really understand what you're trying to get at in the first part. Would you mind going into a bit more detail? I'm quite new to this. Sorry again.
$endgroup$
– user494405
Dec 4 '18 at 23:51
$begingroup$
Ah, I think I /may/ understand. Are you comparing the terms to other terms which obviously converge? Seeing as they are less than the convergent integrals that you've compared them to, the integral must be convergent?
$endgroup$
– user494405
Dec 4 '18 at 23:54
$begingroup$
@user494405 that's exactly right
$endgroup$
– qbert
Dec 5 '18 at 0:09
$begingroup$
Sorry to keep asking questions, but would you mind sharing how you decided on those particular comparisons? What is the general thought process which led you to choose them? Thanks again.
$endgroup$
– user494405
Dec 5 '18 at 0:13
$begingroup$
there are two important facts about this integral: Near $0$, $e^{-x}$ is pretty much $1$, and $sqrt{1}{sqrt{x}}$ is integrable, so find a way to establish a bound in terms of $frac{1}{sqrt{x}}$, and then for large $x$, bounding by the very convergent $e^{-x}$ seems clear enough. Breaking an integral into a part near a finite point where you are worried about it blowing up, and near infinity is common
$endgroup$
– qbert
Dec 5 '18 at 0:16
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026351%2fhow-to-determine-if-an-improper-integral-converges-or-diverges%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Finding an antiderivative is going to be difficult. Seeing if it converges will not,
$$
int_0^infty frac{e^{-x}}{sqrt{x}}mathrm dx=
int_0^1 frac{e^{-x}}{sqrt{x}}mathrm dx+int_1^infty frac{e^{-x}}{sqrt{x}}mathrm dx
$$
Now, for $0leq xleq 1$, we have
$$
frac{e^{-x}}{sqrt{x}}leq frac{1}{sqrt{x}}
$$
and for $1leq x$, we have
$$
frac{e^{-x}}{sqrt{x}}leq e^{-x}
$$
Alternatively, enforce the substitution $u=sqrt{x}$, then your integral is
$$
int_0^infty frac{e^{-x}}{sqrt{x}}mathrm dx=
2int_0^infty e^{-u^2}mathrm du=sqrt{pi}
$$
answered Dec 4 '18 at 23:42
qbertqbert
22.1k32560
$endgroup$
$begingroup$
Thanks for your reponse. Sorry to be dense, but I don't really understand what you're trying to get at in the first part. Would you mind going into a bit more detail? I'm quite new to this. Sorry again.
$endgroup$
– user494405
Dec 4 '18 at 23:51
$begingroup$
Ah, I think I /may/ understand. Are you comparing the terms to other terms which obviously converge? Seeing as they are less than the convergent integrals that you've compared them to, the integral must be convergent?
$endgroup$
– user494405
Dec 4 '18 at 23:54
$begingroup$
@user494405 that's exactly right
$endgroup$
– qbert
Dec 5 '18 at 0:09
$begingroup$
Sorry to keep asking questions, but would you mind sharing how you decided on those particular comparisons? What is the general thought process which led you to choose them? Thanks again.
$endgroup$
– user494405
Dec 5 '18 at 0:13
$begingroup$
there are two important facts about this integral: Near $0$, $e^{-x}$ is pretty much $1$, and $sqrt{1}{sqrt{x}}$ is integrable, so find a way to establish a bound in terms of $frac{1}{sqrt{x}}$, and then for large $x$, bounding by the very convergent $e^{-x}$ seems clear enough. Breaking an integral into a part near a finite point where you are worried about it blowing up, and near infinity is common
$endgroup$
– qbert
Dec 5 '18 at 0:16
add a comment |
$begingroup$
Finding an antiderivative is going to be difficult. Seeing if it converges will not,
$$
int_0^infty frac{e^{-x}}{sqrt{x}}mathrm dx=
int_0^1 frac{e^{-x}}{sqrt{x}}mathrm dx+int_1^infty frac{e^{-x}}{sqrt{x}}mathrm dx
$$
Now, for $0leq xleq 1$, we have
$$
frac{e^{-x}}{sqrt{x}}leq frac{1}{sqrt{x}}
$$
and for $1leq x$, we have
$$
frac{e^{-x}}{sqrt{x}}leq e^{-x}
$$
Alternatively, enforce the substitution $u=sqrt{x}$, then your integral is
$$
int_0^infty frac{e^{-x}}{sqrt{x}}mathrm dx=
2int_0^infty e^{-u^2}mathrm du=sqrt{pi}
$$
answered Dec 4 '18 at 23:42
qbertqbert
22.1k32560
$endgroup$
$begingroup$
Thanks for your reponse. Sorry to be dense, but I don't really understand what you're trying to get at in the first part. Would you mind going into a bit more detail? I'm quite new to this. Sorry again.
$endgroup$
– user494405
Dec 4 '18 at 23:51
$begingroup$
Ah, I think I /may/ understand. Are you comparing the terms to other terms which obviously converge? Seeing as they are less than the convergent integrals that you've compared them to, the integral must be convergent?
$endgroup$
– user494405
Dec 4 '18 at 23:54
$begingroup$
@user494405 that's exactly right
$endgroup$
– qbert
Dec 5 '18 at 0:09
$begingroup$
Sorry to keep asking questions, but would you mind sharing how you decided on those particular comparisons? What is the general thought process which led you to choose them? Thanks again.
$endgroup$
– user494405
Dec 5 '18 at 0:13
$begingroup$
there are two important facts about this integral: Near $0$, $e^{-x}$ is pretty much $1$, and $sqrt{1}{sqrt{x}}$ is integrable, so find a way to establish a bound in terms of $frac{1}{sqrt{x}}$, and then for large $x$, bounding by the very convergent $e^{-x}$ seems clear enough. Breaking an integral into a part near a finite point where you are worried about it blowing up, and near infinity is common
$endgroup$
– qbert
Dec 5 '18 at 0:16
add a comment |
$begingroup$
Finding an antiderivative is going to be difficult. Seeing if it converges will not,
$$
int_0^infty frac{e^{-x}}{sqrt{x}}mathrm dx=
int_0^1 frac{e^{-x}}{sqrt{x}}mathrm dx+int_1^infty frac{e^{-x}}{sqrt{x}}mathrm dx
$$
Now, for $0leq xleq 1$, we have
$$
frac{e^{-x}}{sqrt{x}}leq frac{1}{sqrt{x}}
$$
and for $1leq x$, we have
$$
frac{e^{-x}}{sqrt{x}}leq e^{-x}
$$
Alternatively, enforce the substitution $u=sqrt{x}$, then your integral is
$$
int_0^infty frac{e^{-x}}{sqrt{x}}mathrm dx=
2int_0^infty e^{-u^2}mathrm du=sqrt{pi}
$$
answered Dec 4 '18 at 23:42
qbertqbert
22.1k32560
$endgroup$
Finding an antiderivative is going to be difficult. Seeing if it converges will not,
$$
int_0^infty frac{e^{-x}}{sqrt{x}}mathrm dx=
int_0^1 frac{e^{-x}}{sqrt{x}}mathrm dx+int_1^infty frac{e^{-x}}{sqrt{x}}mathrm dx
$$
Now, for $0leq xleq 1$, we have
$$
frac{e^{-x}}{sqrt{x}}leq frac{1}{sqrt{x}}
$$
and for $1leq x$, we have
$$
frac{e^{-x}}{sqrt{x}}leq e^{-x}
$$
Alternatively, enforce the substitution $u=sqrt{x}$, then your integral is
$$
int_0^infty frac{e^{-x}}{sqrt{x}}mathrm dx=
2int_0^infty e^{-u^2}mathrm du=sqrt{pi}
$$
answered Dec 4 '18 at 23:42
qbertqbert
22.1k32560
answered Dec 4 '18 at 23:42
qbertqbert
22.1k32560
answered Dec 4 '18 at 23:42
qbertqbert
22.1k32560
answered Dec 4 '18 at 23:42
qbertqbert
22.1k32560
22.1k32560
$begingroup$
Thanks for your reponse. Sorry to be dense, but I don't really understand what you're trying to get at in the first part. Would you mind going into a bit more detail? I'm quite new to this. Sorry again.
$endgroup$
– user494405
Dec 4 '18 at 23:51
$begingroup$
Ah, I think I /may/ understand. Are you comparing the terms to other terms which obviously converge? Seeing as they are less than the convergent integrals that you've compared them to, the integral must be convergent?
$endgroup$
– user494405
Dec 4 '18 at 23:54
$begingroup$
@user494405 that's exactly right
$endgroup$
– qbert
Dec 5 '18 at 0:09
$begingroup$
Sorry to keep asking questions, but would you mind sharing how you decided on those particular comparisons? What is the general thought process which led you to choose them? Thanks again.
$endgroup$
– user494405
Dec 5 '18 at 0:13
$begingroup$
there are two important facts about this integral: Near $0$, $e^{-x}$ is pretty much $1$, and $sqrt{1}{sqrt{x}}$ is integrable, so find a way to establish a bound in terms of $frac{1}{sqrt{x}}$, and then for large $x$, bounding by the very convergent $e^{-x}$ seems clear enough. Breaking an integral into a part near a finite point where you are worried about it blowing up, and near infinity is common
$endgroup$
– qbert
Dec 5 '18 at 0:16
add a comment |
$begingroup$
Thanks for your reponse. Sorry to be dense, but I don't really understand what you're trying to get at in the first part. Would you mind going into a bit more detail? I'm quite new to this. Sorry again.
$endgroup$
– user494405
Dec 4 '18 at 23:51
$begingroup$
Ah, I think I /may/ understand. Are you comparing the terms to other terms which obviously converge? Seeing as they are less than the convergent integrals that you've compared them to, the integral must be convergent?
$endgroup$
– user494405
Dec 4 '18 at 23:54
$begingroup$
@user494405 that's exactly right
$endgroup$
– qbert
Dec 5 '18 at 0:09
$begingroup$
Sorry to keep asking questions, but would you mind sharing how you decided on those particular comparisons? What is the general thought process which led you to choose them? Thanks again.
$endgroup$
– user494405
Dec 5 '18 at 0:13
$begingroup$
there are two important facts about this integral: Near $0$, $e^{-x}$ is pretty much $1$, and $sqrt{1}{sqrt{x}}$ is integrable, so find a way to establish a bound in terms of $frac{1}{sqrt{x}}$, and then for large $x$, bounding by the very convergent $e^{-x}$ seems clear enough. Breaking an integral into a part near a finite point where you are worried about it blowing up, and near infinity is common
$endgroup$
– qbert
Dec 5 '18 at 0:16
$begingroup$
Thanks for your reponse. Sorry to be dense, but I don't really understand what you're trying to get at in the first part. Would you mind going into a bit more detail? I'm quite new to this. Sorry again.
$endgroup$
– user494405
Dec 4 '18 at 23:51
$begingroup$
Thanks for your reponse. Sorry to be dense, but I don't really understand what you're trying to get at in the first part. Would you mind going into a bit more detail? I'm quite new to this. Sorry again.
$endgroup$
– user494405
Dec 4 '18 at 23:51
$begingroup$
Ah, I think I /may/ understand. Are you comparing the terms to other terms which obviously converge? Seeing as they are less than the convergent integrals that you've compared them to, the integral must be convergent?
$endgroup$
– user494405
Dec 4 '18 at 23:54
$begingroup$
Ah, I think I /may/ understand. Are you comparing the terms to other terms which obviously converge? Seeing as they are less than the convergent integrals that you've compared them to, the integral must be convergent?
$endgroup$
– user494405
Dec 4 '18 at 23:54
$begingroup$
@user494405 that's exactly right
$endgroup$
– qbert
Dec 5 '18 at 0:09
$begingroup$
@user494405 that's exactly right
$endgroup$
– qbert
Dec 5 '18 at 0:09
$begingroup$
Sorry to keep asking questions, but would you mind sharing how you decided on those particular comparisons? What is the general thought process which led you to choose them? Thanks again.
$endgroup$
– user494405
Dec 5 '18 at 0:13
$begingroup$
Sorry to keep asking questions, but would you mind sharing how you decided on those particular comparisons? What is the general thought process which led you to choose them? Thanks again.
$endgroup$
– user494405
Dec 5 '18 at 0:13
$begingroup$
there are two important facts about this integral: Near $0$, $e^{-x}$ is pretty much $1$, and $sqrt{1}{sqrt{x}}$ is integrable, so find a way to establish a bound in terms of $frac{1}{sqrt{x}}$, and then for large $x$, bounding by the very convergent $e^{-x}$ seems clear enough. Breaking an integral into a part near a finite point where you are worried about it blowing up, and near infinity is common
$endgroup$
– qbert
Dec 5 '18 at 0:16
$begingroup$
there are two important facts about this integral: Near $0$, $e^{-x}$ is pretty much $1$, and $sqrt{1}{sqrt{x}}$ is integrable, so find a way to establish a bound in terms of $frac{1}{sqrt{x}}$, and then for large $x$, bounding by the very convergent $e^{-x}$ seems clear enough. Breaking an integral into a part near a finite point where you are worried about it blowing up, and near infinity is common
$endgroup$
– qbert
Dec 5 '18 at 0:16
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026351%2fhow-to-determine-if-an-improper-integral-converges-or-diverges%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
a) Do you really want to mix $x$ and $t$? b) If you just want to know some estimates, then estimate this is reasonably close to $int_0^1 t^{-1/2} , dt$.
$endgroup$
– T. Bongers
Dec 4 '18 at 23:38
1
$begingroup$
You have an $e^{-x}$ in your integrand. Is that $x$ meant to be a $t$? If not, the integrand is much more trivial to deal with since it'd be a constant w.r.t. $t$.
$endgroup$
– Eevee Trainer
Dec 4 '18 at 23:38
$begingroup$
@T.Bongers Apologies, had a brain fart. Yes, I meant to have everything be the same variable; fixed now.
$endgroup$
– user494405
Dec 4 '18 at 23:40