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This is something I came up with off the top of my head, not a homework problem, so it might not have a simple solution. $$\int_1^\infty ln(\frac{t-x}{t})dt$$ Where $x$ is a positive real number. I was able to find the antiderivative: $$t\ln(\frac{t-x}{t})-x\ln(t-x)$$ Which was pretty straightforward but got caught up trying to evaluate it at infinity. It seems like the first term should be either finite or zero, but the second term is definitely infinite. Yet Wolfram says it converges, so what is the proper way of evaluating it?

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    Doesn't seem to converge near infinity. – 2017-02-16
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    @ZaidAlyafeai I would've assumed the same thing but Wolfram's never failed me: http://www.wolframalpha.com/input/?i=integrate+ln((t-x)%2Ft)+wrt+t+from+1+to+infty – 2017-02-16
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    I don't expect the integral to converge. For big t, the integrand function is $-x/t$ to first order. – 2017-02-16
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    Never trust a software. – 2017-02-16

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\begin{align*} t\ln\left(\frac{t-x}{t}\right)-x\ln(t-x)&=\ln\left(\frac{(t-x)^t}{t^t}\right)-\ln\left((t-x)^x\right)\\ &=\ln\left(\frac{(t-x)^{(t-x)}}{t^t}\right)\\ \end{align*}

For $x>0$, as $t\rightarrow\infty$, this diverges to $-\infty$. For $x<0$, as $t\rightarrow-\infty$, this diverges to $\infty$.

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    As t grows wouldn't the denominator approach $t^t$? And wouldn't that make the limit zero? – 2017-02-16
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    It would make the fraction equal to 0, but there is also natural ligarithm enclosing it, so it would translate into $\ln(0)$ which tends to infinity – 2017-02-16
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    @KKZiomek I meant numerator when I said denominator. But in the limit of very large t, it seems like the argument of the logarithm would be $t^tt^{-t}=t^0=1$ and $\ln(1)=0$ – 2017-02-16
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    Don't forget there is x. What if x is equal to infinity also? Then we get an indeterminate form. – 2017-02-16
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    @KKZiomek A fair point, but this is part of a broader problem I was trying to solve, in the context of which x is finite. I should've mentioned that in the question so I just editied it. – 2017-02-16