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# Integration Challange

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

Note by Nishu Sharma
2 years, 1 month ago

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Can anyone explain this https://brilliant.org/discussions/thread/extremely-weird-integration/?ref_id=946409 · 1 year, 8 months 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, $$\displaystyle \frac{{I}_{1}}{{I}_{2}} = {e}^{{x}^{2}/4}$$ · 2 years, 1 month ago

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Kartik there was no need to uneccasarily bring the error function into play here. · 2 years, 1 month ago

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Yes, you can convert one integral directly into the other without erf(). · 2 years, 1 month ago

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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 ! · 2 years, 1 month ago

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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. @Nishu sharma · 2 years ago

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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 ? · 2 years ago

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Yeah, I agree. That was a mistake! · 2 years, 1 month ago

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Nicely done , but there is more Smarter approach for this .. Think about it Kartik sharma · 2 years, 1 month ago

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Also, never miss out $$dz$$ inside integral. It is a bad habit. · 2 years, 1 month ago

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Yeah. But it is tedious. :P · 2 years, 1 month ago

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The answer is:

$\displaystyle \large \large e^{\frac{x^2}{4}}$ · 2 years, 1 month ago

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Yes ! But I gave this as a Challange , So Everyone should post thier approaches .. Like karthik did ! · 2 years, 1 month ago

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