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Does anyone have any advice or help on how to analytically solve the following problem?

Prove that the function $$ \operatorname{f}\left(r\right) = \int_{0}^{\infty}\operatorname{J}_{0}\left(rx\right)\, \mathrm e^{-ax^{2}}\cos\left(x\right)x\,\mathrm dx\,,\quad r \geq 0 $$ is positive beyond a certain value for $a$ positive. Show that the exact threshold for $a$ is the one needed to ensure that it is positive at the origin.

Some notes:

• It is easy to calculate the exact value of $a$ beyond which $\operatorname{f}\left(0\right)$ is positive. Also, $\operatorname{f}\left(r\right) \to 0$ as $r \to \infty$.
• So if I can prove that $\operatorname{f}$ cannot have any zeros (or local minima) it would prevent it from becoming negative.
• I am not sure though if looking at the critical points of $\operatorname{f}$ is the right way to go.

Thanks in advance for any help :) Even help in gauging how easy/difficult this question is is welcome.

Does anyone have any advice or help on how to analytically solve the following problem?

Prove that the function $$ \operatorname{f}\left(r\right) = \int_{0}^{\infty}\operatorname{J}_{0}\left(rx\right)\, \mathrm e^{-ax^{2}}\cos\left(x\right)x\,\mathrm dx\,,\quad r \geq 0 $$ is positive beyond a certain value for $a$ positive, where the exact threshold for $a$ is the one needed to ensure that it is positive at the origin.

Some notes:

• It is easy to calculate the exact value of $a$ beyond which $\operatorname{f}\left(0\right)$ is positive. Also, $\operatorname{f}\left(r\right) \to 0$ as $r \to \infty$.
• So if I can prove that $\operatorname{f}$ cannot have any zeros (or local minima) it would prevent it from becoming negative.
• I am not sure though if looking at the critical points of $\operatorname{f}$ is the right way to go.

Thanks in advance for any help :) Even help in gauging how easy/difficult this question is is welcome.

Does anyone have any advice or help on how to tackle the following problem?

Prove that the function $$ \operatorname{f}\left(r\right) = \int_{0}^{\infty}\operatorname{J}_{0}\left(rx\right)\, \mathrm e^{-ax^{2}}\cos\left(x\right)x\,\mathrm dx\,,\quad r \geq 0 $$ is positive beyond a certain value for $a$ positive. Would it be possible to rigorously show that the exact threshold for $a$ is the one needed to ensure that it is positive at the origin?

Some notes:

• It is easy to calculate the exact value of $a$ beyond which $\operatorname{f}\left(0\right)$ is positive. Also, $\operatorname{f}\left(r\right) \to 0$ as $r \to \infty$.
• So if I can prove that $\operatorname{f}$ cannot have any zeros (or local minima) it would prevent it from becoming negative.
• I am not sure though if looking at the critical points of $\operatorname{f}$ is the right way to go.

Thanks in advance for any help :) Even help in gauging how easy/difficult this question is is welcome.

Does anyone have any advice or help on how to analytically solve the following problem?

Prove that the function $$ \operatorname{f}\left(r\right) = \int_{0}^{\infty}\operatorname{J}_{0}\left(rx\right)\, \mathrm e^{-ax^{2}}\cos\left(x\right)x\,\mathrm dx\,,\quad r \geq 0 $$ is positive beyond a certain value for $a$ positive. Show that the exact threshold for $a$ is the one needed to ensure that it is positive at the origin.

Some notes:

• It is easy to calculate the exact value of $a$ beyond which $\operatorname{f}\left(0\right)$ is positive. Also, $\operatorname{f}\left(r\right) \to 0$ as $r \to \infty$.
• So if I can prove that $\operatorname{f}$ cannot have any zeros (or local minima) it would prevent it from becoming negative.
• I am not sure though if looking at the critical points of $\operatorname{f}$ is the right way to go.

Thanks in advance for any help :) Even help in gauging how easy/difficult this question is is welcome.

Does anyone have any advice or help on how to tackle the following problem?

Show that the function $$ \operatorname{f}\left(r\right) = \int_{0}^{\infty}\operatorname{J}_{0}\left(rx\right)\, \mathrm e^{-ax^{2}}\cos\left(x\right)x\,\mathrm dx\,,\quad r \geq 0 $$ is positive beyond a certain value for $a$ positive. Would it be possible to rigorously show that the exact threshold for $a$ is the one needed to ensure that it is positive at the origin?

Some notes:

• It is easy to calculate the exact value of $a$ beyond which $\operatorname{f}\left(0\right)$ is positive. Also, $\operatorname{f}\left(r\right) \to 0$ as $r \to \infty$.
• So if I can prove that $\operatorname{f}$ cannot have any zeros (or local minima) it would prevent it from becoming negative.
• I am not sure though if looking at the critical points of $\operatorname{f}$ is the right way to go.

Thanks in advance for any help :) Even help in gauging how easy/difficult this question is is welcome.

Does anyone have any advice or help on how to tackle the following problem?

Prove that the function $$ \operatorname{f}\left(r\right) = \int_{0}^{\infty}\operatorname{J}_{0}\left(rx\right)\, \mathrm e^{-ax^{2}}\cos\left(x\right)x\,\mathrm dx\,,\quad r \geq 0 $$ is positive beyond a certain value for $a$ positive. Would it be possible to rigorously show that the exact threshold for $a$ is the one needed to ensure that it is positive at the origin?

Some notes:

• It is easy to calculate the exact value of $a$ beyond which $\operatorname{f}\left(0\right)$ is positive. Also, $\operatorname{f}\left(r\right) \to 0$ as $r \to \infty$.
• So if I can prove that $\operatorname{f}$ cannot have any zeros (or local minima) it would prevent it from becoming negative.
• I am not sure though if looking at the critical points of $\operatorname{f}$ is the right way to go.

Thanks in advance for any help :) Even help in gauging how easy/difficult this question is is welcome.