# $I_n=\int_0^1{xe^{-nx^2}}$. Prove that $I_n=\frac{1}{2n}(1-\frac{1}{e^n})$

$$I_n=\int_0^1{xe^{-nx^2}}dx$$ 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.

• 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. – Radu Gabriel Jan 30 '20 at 19:15
• Just substitute $nx^2=t$. – Zacky Jan 30 '20 at 19:17
• Yeah I can't believe I did not see it. I think I can solve the exercise now. – Radu Gabriel Jan 30 '20 at 19:18

$$I_n=\int_0^1{xe^{-nx^2}}dx$$
$$I_n=-\frac {1}{2n}\int_0^1-2n{xe^{-nx^2}}dx$$ $$I_n=-\frac {1}{2n}\int_0^1\left (e^{-nx^2}\right )'dx$$ 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 $$I_n=\frac {1}{2n} \left (1-e^{-n}\right )$$