Let \[\displaystyle{{ I }_{ 1 }=\int _{ 0 }^{ x }{ { e }^{ zx }{ e }^{ -{ z }^{ 2 } } } dz\\ { I }_{ 2 }=\int _{ 0 }^{ x }{ { e }^{ -\cfrac { { z }^{ 2 } }{ 4 } }dz } \\ \boxed { \cfrac { { I }_{ 1 } }{ { I }_{ 2 } } =? } }\]

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TopNewestCan anyone explain this https://brilliant.org/discussions/thread/extremely-weird-integration/?ref_id=946409 – 柯 南 · 2 years ago

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Well, it's very easy. And Raghav has given exactly the correct answer considering it to be very easy.

\(\displaystyle {I}_{1} = \int_{0}^{x}{{e}^{-{z}^{2}+zx}}\)

Completing the square,

\(\displaystyle {I}_{1} = \int_{0}^{x}{{e}^{{x}^{2}/4}{e}^{-{z}^{2}+zx+{x}^{2}/4}}\)

\(\displaystyle {I}_{1} = {e}^{{x}^{2}/4}\int_{0}^{x}{{e}^{-{(z-x/2)}^{2}}}\)

Using u-substitution and a little bashing up, we get

\(\displaystyle {I}_{1} = {e}^{{x}^{2}/4}\int_{-x/2}^{x/2}{{e}^{-{u}^{2}}}\)

And hence, \(\displaystyle {I}_{1} = 2{e}^{{x}^{2}/4}\int_{0}^{x/2}{{e}^{-{u}^{2}}}\)

By using error function now,

\(\displaystyle {I}_{1} = {e}^{{x}^{2}/4}\sqrt{\pi}\text{erf}(x/2)\)

Now, \(\displaystyle {I}_{2} = \int_{0}^{x}{{e}^{-{(z/2)}^{2}}}\)

Using v-substitution and some 'using of facts',

\(\displaystyle {I}_{2} = 2\int_{0}^{x/2}{{e}^{-{v}^{2}}}\)

\(\displaystyle {I}_{2} = \sqrt{\pi}\text{erf}(x/2)\)

Therefore, \(\displaystyle \frac{{I}_{1}}{{I}_{2}} = {e}^{{x}^{2}/4}\) – Kartik Sharma · 2 years, 4 months ago

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– Ronak Agarwal · 2 years, 4 months ago

Kartik there was no need to uneccasarily bring the error function into play here.Log in to reply

– Raghav Vaidyanathan · 2 years, 4 months ago

Yes, you can convert one integral directly into the other without erf().Log in to reply

– Nishu Sharma · 2 years, 4 months ago

Hi Ronak ... I have solved Your Physics questions , I found all of them very Interesting . I really much impressed from your Problem making skills . I really appreciate awesomeness of your Problems ! Great work .. ! I'am regret that I joined brilliant very late , My relative brother recommended this site to me recently ... There are so many good questions of maths and physics , and I found that :: You and Jatin yadav and Deepanshu and David Mattingly and Josh silverman Posted so many self made questions .. I'am really amazed with your skills !Log in to reply

@Nishu sharma – Ronak Agarwal · 2 years, 4 months ago

I know it's pretty late to type this but thanks for the appreciation, I see you are from jaipur also, in which area of jaipur are you from, I am from jaipur too.Log in to reply

– Nishu Sharma · 2 years, 4 months ago

Hi , Actually My home is at jaipur , but I live in kota alone , near by to my aunt , My family lives in Jaipur. My home in backyard of Vidhan-sabha .. where do you live ?Log in to reply

– Kartik Sharma · 2 years, 4 months ago

Yeah, I agree. That was a mistake!Log in to reply

– Nishu Sharma · 2 years, 4 months ago

Nicely done , but there is more Smarter approach for this .. Think about it Kartik sharmaLog in to reply

– Raghav Vaidyanathan · 2 years, 4 months ago

Also, never miss out \(dz\) inside integral. It is a bad habit.Log in to reply

– Kartik Sharma · 2 years, 4 months ago

Yeah. But it is tedious. :PLog in to reply

@Nishu sharma

The answer is:

\[\displaystyle \large \large e^{\frac{x^2}{4}}\] – Raghav Vaidyanathan · 2 years, 4 months ago

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– Nishu Sharma · 2 years, 4 months ago

Yes ! But I gave this as a Challange , So Everyone should post thier approaches .. Like karthik did !Log in to reply

@Brian Charlesworth @Kunal Joshi @Ronak Agarwal @Raghav Vaidyanathan @Calvin Lin – Nishu Sharma · 2 years, 4 months ago

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