Evaluate $\lim_{x\to\infty }x\int_1^{x}\frac{e^t}{t}dt-e^x$ 
Question: Mention (True or False) Let $f:[1,\infty)\to\mathbb{R}$ and $f(x)=x\int_1^{x}\frac{e^t}{t}dt-e^x$ then f(x) is an increasing function and $\lim_{x\to \infty}f(x)\to\infty.$

Differentiating $f(x)$ we can conclude that $f(x)$ is increasing. Using $e^t=1+t+\frac{t^2}{2!}+\dots$
Then $$f(x)=x\int_1^x(\frac{1}{t}+1+\frac{t}{2!}+\frac{t^2}{3!}+\dots)dt-e^x=x(\ln(t)+t+\frac{t^2}{2(2!)}+\dots)_1^x-e^x=x(\ln(x)+x+\frac{x^2}{2(2!)}+\dots-1-\frac{1}{2(2!)}-\dots)-e^x\\$$
It seems that for $x\to\infty$ we will have $f(x)\to-\infty$ because of the rate of growth of the exponent .
The argument that I have mentioned is merely an intuition, this does not seems correct . Thanks.
 A: This is a way you could formaly express your intuition:
$$f(x)=x\int_1^x\Big(\frac{1}{t}+\sum_{n=1}^\infty\frac{t^{n-1}}{n!}\Big)dt-e^x=x\int_1^x\frac{1}{t}dt+x\int_1^x\sum_{n=1}^\infty\frac{t^{n-1}}{n!}dt-e^x=$$
$$=x\ln x+x\int_1^x\sum_{n=1}^\infty\frac{t^{n-1}}{n!}dt-e^x=x\ln x+x\sum_{n=1}^\infty\int_1^x\frac{t^{n-1}}{n!}dt-e^x=$$
$$=x\ln x+x\sum_{n=1}^\infty\frac{x^n-1}{n\cdot n!}-e^x.$$
The interchange of the integral and the sum is permitted by Fubini's theorem since all the summands are always positive. Now you just need to show that this final expression reaches infinity. You didn't include that in your intuitive part, but I will include it here too:
$$f(x)=x\ln x+x\sum_{n=1}^\infty\frac{x^n-1}{n\cdot n!}-e^x=x\ln x+x\sum_{n=1}^\infty\frac{x^n}{n\cdot n!}-x\sum_{n=1}^\infty\frac{1}{n\cdot n!}-e^x>$$
$$>x\ln x+x\sum_{n=1}^\infty\frac{x^n}{(n+1)!}-x\sum_{n=1}^\infty\frac{1}{n!}-e^x=x\ln x+(e^x-x-1)-x\cdot e-e^x=$$
$$=x(\ln x-1-e)-1>x\rightarrow\infty.$$
The second to the last inequality holds whenever $\ln x>2+e+\frac{1}{x}$, which happens for all $x$ sufficiently large (at about $113$).
A: Let us use the fundamental theorem of calculus and the mean value theorem, alone with some very rough estimates.
If we set $f(x)=x\int_1^{x}\frac{e^t}{t}dt-e^x$, then the product rule and the fundamental theorem of calculus yield
$$f'(x)=\int_1^{x}\frac{e^t}{t}dt$$
and if we can get rough bounds for this quantity, we can get rough bounds for $f(x)$.
If $g(x)=e^x/x$, then $g'(x)=\frac{(x-1)e^x}{x^2}$, and thus $g(x)$ is increasing on $(1,\infty)$.  In particular, if $x\geq 1$, then $g(x)\geq g(1)=e$.    Therefore if $x\geq 1, f'(x)=\int_1^x g(t)dt \geq (x-1)e$.  Consequently, if $x>1$, $$f(x)=f(1)+\int_1^{x}f'(t)dt\geq -e+\int_1^x e(t-1)dt=e(x^2/2-x-(1/2)).$$
This polynomial lower bound is obviously not bounded above, and thus $f(x)$ must approach infinity.
A: Here is a different way to prove that $f(x)$ approaches infinity. This one doesn't have any series which could confuse you. Your intuition "failed" (tricked you) because you forgot to multiply your brackets with that $x$ outside. If you compared the coefficients after that, you would guess the correct answer. For all $x>2$ we have:
$$f(x)=x\int_1^x\frac{e^t}{t}dt-e^x>x\int_1^{\frac{x}{2}}\frac{e^t}{\frac{x}{2}}dt+x\int_{\frac{x}{2}}^x\frac{e^t}{x}dt-e^x=$$
$$=x\int_1^{\frac{x}{2}}\frac{2e^t}{x}dt+x\int_{\frac{x}{2}}^x\frac{e^t}{x}dt-e^x=x\int_1^{\frac{x}{2}}\frac{e^t}{x}dt+x\int_1^{\frac{x}{2}}\frac{e^t}{x}dt+x\int_{\frac{x}{2}}^x\frac{e^t}{x}dt-e^x=$$
$$=x\int_1^{\frac{x}{2}}\frac{e^t}{x}dt+x\int_{1}^x\frac{e^t}{x}dt-e^x=\int_1^{\frac{x}{2}}e^tdt+\int_{1}^xe^tdt-e^x=$$
$$=e^{\frac{x}{2}}-e+e^x-e-e^x=e^{\frac{x}{2}}-2e.$$
The last expression obviouslt approaches $\infty$. The spirit of this solution was based on
$$x\int_1^x\frac{e^t}{t}dt>x\int_1^x\frac{e^t}{x}dt,$$
but it wasn't enough, so I split the integral into $2$ parts and did this trick on both of them. To prove that $f$ is increasing, we can simply take its derivative:
$$f'(x)=\int_1^x\frac{e^t}{t}dt+x\frac{e^x}{x}-e^x=\int_1^x\frac{e^t}{t}dt>0.$$
Then the third way to prove that it approaches infinity would be to calculate, once again, its second derivative:
$$f''(x)=\frac{d}{dx}f'(x)=\frac{e^x}{x}>0,$$
and show that $f$ is convex. Any convex function that is increasing must approach infinity.
