$$I_n=\int_0^1{xe^{-nx^2}}dx$$ Prove that $I_n=\frac{1}{2n}(1-\frac{1}{e^n})$. I also know that $I_{n+1}-I_n \le 0$. I had to prove this previously and this is how I did it (I am putting this in here because I believe I've done something wrong and this is why I can't get a useful relationship between $I_n$ and $I_{n+1}$) $$I_{n+1}-I_n=\int_0^1{\frac{x}{e^{(n+1)x^2}}-\frac{x}{e^{nx^2}}}dx$$ $$I_{n+1}-I_n=\int_0^1{\frac{x}{e^{nx^2}\cdot e^{x^2}}-\frac{x}{e^{nx^2}}}dx$$ $$I_{n+1}-I_n=\int_0^1{\frac{x}{e^{nx^2}}}\bigg(\frac{1}{e^{x^2}}-1\bigg)dx$$ Now the first part of the integral is clearly positive for $x \in (0,1)$ then I took the part between the parantheses and observed it's negative for $x\in(0,1)$ so the inequality is true.

I figured the second part of the exercise(the one the answer is about) can be solved using mathematical induction and I tried to calculate $I_{n+1}$ based on $I_n$ from the previous relationship: $$I_{n+1}=I_n\int_0^1\bigg(\frac{1}{e^{x^2}}-1\bigg)-I_n$$ $$I_{n+1}=I_n\biggr(\int_0^1\bigg(\frac{1}{e^{x^2}}-1\bigg)+1\biggr)$$
Now, I don't think that $\int_0^1{\big(\frac{1}{e^{x^2}}-1\big)}dx$ can be solved so this leads me to believe that I have made a very basic algebraic while proving the inequality but I just can't see it, went over it countless times.

  • $\begingroup$ Actually, as I was reading this it hit me that $I_n$ can be calculated with u-substitution. I'll give it a try now. $\endgroup$ – Radu Gabriel Jan 30 '20 at 19:15
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    $\begingroup$ Just substitute $nx^2=t$. $\endgroup$ – Zacky Jan 30 '20 at 19:17
  • $\begingroup$ Yeah I can't believe I did not see it. I think I can solve the exercise now. $\endgroup$ – Radu Gabriel Jan 30 '20 at 19:18

I don't know how to type but you will be able to understand this(https://i.stack.imgur.com/DEit4.jpg)

  • $\begingroup$ math.meta.stackexchange.com/questions/5020/… $\endgroup$ – StubbornAtom Jan 30 '20 at 19:31
  • $\begingroup$ Thanks for your details $\endgroup$ – Akalanka Jan 30 '20 at 19:32
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    $\begingroup$ I appreciate you helping, again. Also, typing with mathjax seems very hard at first but once you get used to it, you'll find it very natural. $\endgroup$ – Radu Gabriel Jan 30 '20 at 20:27
  • $\begingroup$ Thank you i will try $\endgroup$ – Akalanka Jan 31 '20 at 5:15


$$I_n=-\frac {1}{2n}\int_0^1-2n{xe^{-nx^2}}dx$$ $$I_n=-\frac {1}{2n}\int_0^1\left (e^{-nx^2}\right )'dx$$ It's a derivative inside the integral: $$I_n=-\frac {1}{2n} \left |\left (e^{-nx^2}\right ) \right |_0^1$$ $$I_n=-\frac {1}{2n} \left (e^{-n}-1\right ) $$ Finally $$I_n=\frac {1}{2n} \left (1-e^{-n}\right ) $$


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